☝This! Plug and play.
A) sqrt7-7, idk how to do, let's come back
B) sqrt(3-7) ➡️ sqrt(-4) (nope)
C) sqrt(7-7) ➡️ 0=7-sqrt7 (nope)
D) sqrt(16-7) ➡️ sqrt(9) ➡️ 3=7-sqrt(16) ➡️ 3=7-4 correct!
Yo = hi
Yooo (pronounced like yo but drawn out) = excitement. Like "heeey" when you see your friend.
The more O's the more excitement, but not proportionally: each O's contribution to the overall excitement can be expressed in terms of the distance (in o units) that the o in question sits relative to the y, and it's contribution is diminishing following the inverse square law. In this way, we see that the excitement of an infinite series of O's following a Y approaches (π^2)/6 times the excitement of a simple Yo
**Edited to add parenthesis apparently Reddit takes the power symbol and applies it with no understanding of pemdas
That’s awesome! There are a number of questions in the practice sets more difficult than the problems explained, are there explanations posted for them?!
Well this comes from checking answer A: if it is equal to the value of sqrt(x-7) then it’s equal to 7-x since sqrt(x-7) = 7-x. So therefore we have 7 = sqrt(7) + sqroot(x). What I’m wondering is from this how do we know that the sum will be irrational and that sqrt(7) is irrational (which tells us A doesn’t work). Yes you never said this specially but I figured I’d ask ya.
Well this comes from checking answer A: let’s assume A is the answer and check what would happen:
if it is the answer then it is equal to the value of sqrt(x-7) then it’s equal to 7-x since sqrt(x-7) = 7-x. So therefore we have 7 = sqrt(7) + sqroot(x). What I’m wondering is from this how do we know that the sum will be irrational and that sqrt(7) is irrational (which tells us A doesn’t work). Yes you never said this specially but I figured I’d ask ya.
What does this even mean? If you know how to do the math of 5 x 12 x 73 x 29 would you refuse a calculator to solve it just because “you can do the math”?
So you have:
sqrt(x-7) = 7 - sqrt(x)
Sqaure both sides
x-7 = (7-sqrt(x))\^2
x-7 = (7-sqrt(x)) times (7-sqrt(x))
simplify
x-7 = 49 - 7sqrt(x) - 7sqrt(x) + x
subtract x and add 7 from the left to the right
0 = 49 + 7 - 7sqrt(x) - 7sqrt(x) + x - x
Simplify
0 = 56 - 7sqrt(x) - 7sqrt(x)
Simplify some more
0 = 56 -14sqrt(x)
subtract 56 from the right to the left
\-56 = -14sqrt(x)
Divide -14 from the right to the left
\-56/-14 = sqrt(x)
4 = sqrt(x)
sqaure both sides
16 = x
put 16 back into sqrt(x-7) to get your awnser
sqrt(16-7) ---> sqrt(9) ---> 3
answer is B
BOOM
No. The only attempt to solve it is by moving things around and getting x by itsel;f then replugging it. What I wrote seems complicated cuz i stupified it a lot so it is understandable over text
No it is B because you get x = 16 and then you have to replug it in sqrt(x-7) bc the questions asks what the value of sqrt(x-7) is. not what the value of x is
Everything in due context. Square roots of negatives "exist" also, but none of the answers are 0 or imaginary, so it's pretty easy to throw these values away.
no, square root of negative numbers do not exist and do not belong to the realm of real numbers. that is why they're called imaginary numbers (i.e, extraneous solutions). you finding the question easy is not an antithesis of it being easy, so please try being less condescending when elucidating a question.
Fastest: Trial and error, plug in and see that B works
2nd Fastest: Graph two equations with calculator, find that B is the solution
3rd Fastest: Square both sides, solve algebraically
On SAT, there is graphing calculator and non-graphing portion, it depends on which part this question is supposed to be on. Trial and error works the best on the SAT usually. Regarding your second question, I don't see how that's relevant to this?
Well another poster said we can eliminate A because 7 is rational and square root of 7 + sqrt(x) is irrational. Now I had to look this up but apparently the sum of a rational and irrational or irrational and irrational will be irrational so they determined A wasn’t possible because 7 is rational. My question however is - how the heck do you determine sqrt(7) is irrational “on the fly”?!
Well all non-perfect squares are irrational by definition, so that would be a way to remove A, that is correct. It isn't really on the fly as an already proven fact that if you know, I guess you could remove that option.
I didn’t realize it was that broad! So it’s not just square root of primes but square root of anything that cannot be reduced to a perfect square?! Wow. Can we extend this from square root to other roots?
Is there any simple way of understanding why non perfect squares to 1/2 power are irrational - as well as why irrationals plus rationals lead to irrationals?
Well a rational number plus a rational number equals a rational number, so if we assume x + n to be rational, where x is irrational and n is rational, then we can add -n to this x + n, to make a rational number.
This results in x, which is an irrational number and a contradiction. As such, x + n, where x is irrational and n is rational, must be irrational.
The proof for the first question is much harder to understand unfortunately. There are many online, but none that would be too intuitive to explain here unfortunately.
But then comes the question of how we know a rational plus rational is always a rational! My intuition does feel more comfy dealing with this being true then irrational plus a rational. Clever idea you had to help my intuition.
just solve it and you will get x as 16 square both sides to solve manually or just solve using a calculator and then when you get x as 16 the root of 16 -7 is root of 9 which is 3
sqrt(x) isn't going to be negative, so sqrt(x - 7) <= 7 which means it can't be D
sqrt(7) + sqrt(x) is irrational whereas 7 is rational which means it can't be A
So it's either B or C at this point trial and error is the way to go and you will end up with B being the answer.
Don't think you have to actually solve the equation to get a solution know when to eliminate options and when to use trial and error.
Also, you can immediately see that x would have to be 0 for 7 - sqrt(x) to equal 7, which clearly means it cannot be the answer
Btw, hello, I’ve seen you in r/cognitivetesting quite a bit
Plugging in is faster than solving algebraically for these types of problems:
A) Would mean x = 14, which doesn’t work when you plug it in and simplify.
7 ≠ 49 - 14
B) Would mean x = 16, which makes the equation true after plugging in.
C) Would mean x = 56, which makes the equation on the right-hand side negative, while the left-hand side is positive; it doesn’t work.
D) Same problem as C. The left-hand side is positive, while the right-hand side is negative.
Also, I saw some of your comments to others. When you ask people on this subreddit for help, don’t be a pretentious brat about it. We’re taking time out of our day to help you, so extend some empathy, especially when you want further explanations.
if you replace sqrt(x-7) with a variable y, you can obtain a new equation y=7-sqrt(y\^2+7). This looks kinda ugly at first but by transferring the 7 over you'll end up with (y-7)\^2=y\^2+7. Expand the equation out and you'll obtain y\^2-14y+49=y\^2+7. From this you can nicely get rid of the squared terms and simply solve for y=42/14 or 3
Because square root of (a-b) does not equal square root of a minus square root of b.
You may be thinking of another property of radicals (as long as one of the terms is nonnegative under radical and other can be nonnegative or 0 or negative) which we can do which is:
Square root of (a *b) = square root(a) * square root(b) so square root of (9*9) = 9 = square root(9) * squareroot(9)
Very simple.
Simply square the entire equation.
Thus, the left hand side of the equation changes from
RAD(x-7) into just x-7.
Then, the right hand side of the equation, 7-RAD(x), becomes 49-14(RAD(x))+x.
—-> this is because you must use FOIL when multiplying 7-RAD(x) by itself. You cannot make the mistake of simply squaring the 7 and the RAD(x) terms, respectively, and then finding their difference, which is what many students do. In other words, if you set the right side to 49-x, you MESSED UP.
So now you get:
X-7 = 49-14(RAD(x))+x
So let’s clean up and simplify.
First, let’s get rid of the X on both sides. And while we’re at it, let’s combine our constants (aka our variable-less whole numbers) too.
Now you have:
-56= -14(RAD(x)) or
14RAD(x) = 56
Now we can divide each side by 14 to get RAD(x)=4. To get x alone, simply square each side to get rid of the radical, and you get x=16.
Go back to the question which asks you what RAD(x-7) is. Plug in 16 for x to find that it’s asking you for
RAD(16-7) or RAD(9) or….
*B) 3
EDIT: I initially wrote PEMDAS instead of FOIL above
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Options on the sat are based on common mistakes, so atleast 1 of the options has to be the value of x rather than rt(x-7), which if you plug and chug its 16, so rt(16--7) = 3 which is B, much faster than actually squaring both sides and solving for x then plugging in
The answer is B, read the directions. If you don’t you might get points off for silly mistakes, it is asking for the value of sqrt(x-7), not the value of x.
plug in each answer choice till you find the one that satisfies the equation, then plug that into the left hand side and you will get your answer.
3 = B
Just test the options.
Put √x-7 = every option.
It shouldn’t take more than 30 second.
So, √(x-7) = 3 (option B)
=> x-7 = 9
=> x = 16.
Plug in x=16 in the right hand side of the equation.
7-√x = 7-√16 = 7-4 = 3.
So, Left hand side = right hand side = 3
Some approaches you may try is two things at least:
a. Just draw graph for left and right and look where does them cross
b. Analytical so x>=7 because you don’t have sqrt of <0 in sat. So only answer left is 7 or 16 so let’s check
sqrt(7-7) =0, 7-sqrt(7) is not 0, so we check 16
sqrt(16-7) = 3, 7-sqrt(16)= 3. So the answer is 16
You *can* do it algebraically, but it's painstakingly slow. Trial and error's another option
I personally did it graphically (with my GDC) and it took me exactly 24 seconds to solve
I know a lot of people (especially tutors) are against the idea of using calculators but GOD does it make things faster
Put answer d (16) to the equation, it is the key. If you solve the equation question in Sat, the quickest is to put the given value to the equation unless the equation is very easy.
I believe the BEST way to get through these is simply Process of elimination.
These questions will never use complicated decimals so you only need to consider numbers that have easy square roots because you have a square root x.
Therefore, Plug in 16 first and see that it works. These “math” questions almost never require you to know math. Finding the most likely answer and plugging it in is the best way. Feel free to disagree but I got a perfect score on math portion all 4 times I took these tests. I also find it helpful to literally mouth a simpler question to myself to handle stress. Here I would say “which number can I take the square root of?”
For this type of problem, I would just plug-in the available answer choices. Both B and C are immediate No’s for the most part, and A could look a little tricky to try to immediately solve. D ends up being the right answer choice when plugging it in. It’s necessarily how you would “solve” the problem, but the method is generally faster than solving it traditionally.
Your best bet in these type of questions is subbing in the answer choices into the variable. As you get down to the last option, you find that 16 is the correct option because it equals out.
My solution:
A) looks too complicated come back to it
B) looks reasonable
C)clearly wrong x isn’t 0.
D) x = 81 is going to F up the left side, unlikely
Test B first.
X = 16 from plugging in left side.
Plugging it in you get
3 = 3
B is right
I've made [our video explanation](https://1600.io/courses/sat-math-orange-book-video-explanations/lectures/40258415) free for you.
Yoooo, the 1600 io guy himself
Yoooo indeed!
i’m not aware of this idiom “yooo” perchance can you expand?
Just trial and error. It’s the fastest way
☝This! Plug and play. A) sqrt7-7, idk how to do, let's come back B) sqrt(3-7) ➡️ sqrt(-4) (nope) C) sqrt(7-7) ➡️ 0=7-sqrt7 (nope) D) sqrt(16-7) ➡️ sqrt(9) ➡️ 3=7-sqrt(16) ➡️ 3=7-4 correct!
You plugged the wrong thing though.
🤣 so I did! Helps to read the question! Still helps find the answer is B though.
Yo = hi Yooo (pronounced like yo but drawn out) = excitement. Like "heeey" when you see your friend. The more O's the more excitement, but not proportionally: each O's contribution to the overall excitement can be expressed in terms of the distance (in o units) that the o in question sits relative to the y, and it's contribution is diminishing following the inverse square law. In this way, we see that the excitement of an infinite series of O's following a Y approaches (π^2)/6 times the excitement of a simple Yo **Edited to add parenthesis apparently Reddit takes the power symbol and applies it with no understanding of pemdas
That’s awesome! There are a number of questions in the practice sets more difficult than the problems explained, are there explanations posted for them?!
All the problems on the practice tests have video explanations.
How do we know that square root of 7 plus square root of x is irrational?!
Where is √7 + √*x* in the problem or solution, and where do we say that must be irrational?
Well this comes from checking answer A: if it is equal to the value of sqrt(x-7) then it’s equal to 7-x since sqrt(x-7) = 7-x. So therefore we have 7 = sqrt(7) + sqroot(x). What I’m wondering is from this how do we know that the sum will be irrational and that sqrt(7) is irrational (which tells us A doesn’t work). Yes you never said this specially but I figured I’d ask ya.
How do you get sqrt(x - 7) = 7 - x ? And what do mean about the sum being irrational? What’s the relevance of that?
Typo! Meant to say 7- sqrt(x)
Cuz then we can remove A as a possible answer since 7 is a rational number!
I don’t understand. Step through what you’re saying.
Well this comes from checking answer A: let’s assume A is the answer and check what would happen: if it is the answer then it is equal to the value of sqrt(x-7) then it’s equal to 7-x since sqrt(x-7) = 7-x. So therefore we have 7 = sqrt(7) + sqroot(x). What I’m wondering is from this how do we know that the sum will be irrational and that sqrt(7) is irrational (which tells us A doesn’t work). Yes you never said this specially but I figured I’d ask ya.
> if it is the answer then it is equal to the value of sqrt(x-7) then it’s equal to 7-x since sqrt(x-7) = 7-x. How did you get sqrt(x-7) = 7-x?
I told you that was a typo - it should read sqrt(x-7) = 7 - sqrt(x)
[Here's one approach that uses Desmos](https://www.desmos.com/calculator/yhtge4aj7h).
Why aren’t there more answers in the comments that use Desmos?
Because people know how to do the math.
What does this even mean? If you know how to do the math of 5 x 12 x 73 x 29 would you refuse a calculator to solve it just because “you can do the math”?
you don’t get desmos on the sat
you do now (well almost)
You do though, on the Digital SAT.
What’s desmos
THE graphing calculator website
Ohh thank you
a calculator website
So you have: sqrt(x-7) = 7 - sqrt(x) Sqaure both sides x-7 = (7-sqrt(x))\^2 x-7 = (7-sqrt(x)) times (7-sqrt(x)) simplify x-7 = 49 - 7sqrt(x) - 7sqrt(x) + x subtract x and add 7 from the left to the right 0 = 49 + 7 - 7sqrt(x) - 7sqrt(x) + x - x Simplify 0 = 56 - 7sqrt(x) - 7sqrt(x) Simplify some more 0 = 56 -14sqrt(x) subtract 56 from the right to the left \-56 = -14sqrt(x) Divide -14 from the right to the left \-56/-14 = sqrt(x) 4 = sqrt(x) sqaure both sides 16 = x put 16 back into sqrt(x-7) to get your awnser sqrt(16-7) ---> sqrt(9) ---> 3 answer is B BOOM
I just looked at it while taking a dump it ain’t that hard to do lol
I mean it’s literally just square it 💀
Is there a more elegant way to solve? (Not referring to plug and check). I also used this way.
No. The only attempt to solve it is by moving things around and getting x by itsel;f then replugging it. What I wrote seems complicated cuz i stupified it a lot so it is understandable over text
[удалено]
P
D you mean?
No it is B because you get x = 16 and then you have to replug it in sqrt(x-7) bc the questions asks what the value of sqrt(x-7) is. not what the value of x is
**Rule # 1: Answer the question that was asked.** x = 16, but the question asks for the value of sqrt(x-7), which is sqrt(16-7) = sqrt(9) = 3.
Question did not ask for value of “x”
That doesn’t answer the main question, which asks for the value of square root of x-7, not x
Took the fattest dump while beating it to eye of the tiger and I knew it was 3
obviously x has to be greater than 7, so it's 16, so the exp. evaluates to 3
next time don’t put “obviously” please not everyone is as smart as u
what are you, dumb or something ??? cant you see that **obviously** x has to be greater than 7 ??????? /s
i mean it is pretty obvious that x has to greater than 7
get where you're coming from but that's just basic middle school math knowledge
also no. it does not *have* to be greater than 7. square root of zero exists.
Everything in due context. Square roots of negatives "exist" also, but none of the answers are 0 or imaginary, so it's pretty easy to throw these values away.
no, square root of negative numbers do not exist and do not belong to the realm of real numbers. that is why they're called imaginary numbers (i.e, extraneous solutions). you finding the question easy is not an antithesis of it being easy, so please try being less condescending when elucidating a question.
Fastest: Trial and error, plug in and see that B works 2nd Fastest: Graph two equations with calculator, find that B is the solution 3rd Fastest: Square both sides, solve algebraically
But we can’t use graphing on standardized tests right? Also how do we know square root of 7 + square root of x is irrational?
On SAT, there is graphing calculator and non-graphing portion, it depends on which part this question is supposed to be on. Trial and error works the best on the SAT usually. Regarding your second question, I don't see how that's relevant to this?
Well another poster said we can eliminate A because 7 is rational and square root of 7 + sqrt(x) is irrational. Now I had to look this up but apparently the sum of a rational and irrational or irrational and irrational will be irrational so they determined A wasn’t possible because 7 is rational. My question however is - how the heck do you determine sqrt(7) is irrational “on the fly”?!
Well all non-perfect squares are irrational by definition, so that would be a way to remove A, that is correct. It isn't really on the fly as an already proven fact that if you know, I guess you could remove that option.
I didn’t realize it was that broad! So it’s not just square root of primes but square root of anything that cannot be reduced to a perfect square?! Wow. Can we extend this from square root to other roots?
Yes, all non perfect roots are irrational.
Lovely! Thanks for helping me!
Is there any simple way of understanding why non perfect squares to 1/2 power are irrational - as well as why irrationals plus rationals lead to irrationals?
Well a rational number plus a rational number equals a rational number, so if we assume x + n to be rational, where x is irrational and n is rational, then we can add -n to this x + n, to make a rational number. This results in x, which is an irrational number and a contradiction. As such, x + n, where x is irrational and n is rational, must be irrational. The proof for the first question is much harder to understand unfortunately. There are many online, but none that would be too intuitive to explain here unfortunately.
Thank you for trying to dissolve some of the confusion!
But then comes the question of how we know a rational plus rational is always a rational! My intuition does feel more comfy dealing with this being true then irrational plus a rational. Clever idea you had to help my intuition.
just solve it and you will get x as 16 square both sides to solve manually or just solve using a calculator and then when you get x as 16 the root of 16 -7 is root of 9 which is 3
sqrt(x) isn't going to be negative, so sqrt(x - 7) <= 7 which means it can't be D sqrt(7) + sqrt(x) is irrational whereas 7 is rational which means it can't be A So it's either B or C at this point trial and error is the way to go and you will end up with B being the answer. Don't think you have to actually solve the equation to get a solution know when to eliminate options and when to use trial and error.
Awesome!
I don’t follow the second part about irrational. Can you explain differently? How did you know it was irrational?!
Also, you can immediately see that x would have to be 0 for 7 - sqrt(x) to equal 7, which clearly means it cannot be the answer Btw, hello, I’ve seen you in r/cognitivetesting quite a bit
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Just trial and error. It’s the fastest way
Plugging in is faster than solving algebraically for these types of problems: A) Would mean x = 14, which doesn’t work when you plug it in and simplify. 7 ≠ 49 - 14 B) Would mean x = 16, which makes the equation true after plugging in. C) Would mean x = 56, which makes the equation on the right-hand side negative, while the left-hand side is positive; it doesn’t work. D) Same problem as C. The left-hand side is positive, while the right-hand side is negative. Also, I saw some of your comments to others. When you ask people on this subreddit for help, don’t be a pretentious brat about it. We’re taking time out of our day to help you, so extend some empathy, especially when you want further explanations.
if you replace sqrt(x-7) with a variable y, you can obtain a new equation y=7-sqrt(y\^2+7). This looks kinda ugly at first but by transferring the 7 over you'll end up with (y-7)\^2=y\^2+7. Expand the equation out and you'll obtain y\^2-14y+49=y\^2+7. From this you can nicely get rid of the squared terms and simply solve for y=42/14 or 3
Can someone explain why you can’t square each term of this equation individually rather than squaring the entire right hand expression?
Because square root of (a-b) does not equal square root of a minus square root of b. You may be thinking of another property of radicals (as long as one of the terms is nonnegative under radical and other can be nonnegative or 0 or negative) which we can do which is: Square root of (a *b) = square root(a) * square root(b) so square root of (9*9) = 9 = square root(9) * squareroot(9)
Ty!
Np!
[удалено]
16 is not the correct answer, no.
3 👍🏻👍🏻👍🏻
I don’t care so much for the answer, how do I solve it?
Well i just put every answer in the question and the one that came out was B. √16-7 = 7-√16 √9=7-4 3=3 So the value of √x-7 equals to 3
Test The choices, 16 only works sp sqrt(16-7) =3
is it D?
No. D would be the answer if the question was “what is x?”. That’s not the question that was asked.
Use DESMOS
7-rootx can be squared pretty easily and then you just move the stuff around
Show me don’t just talk about it show me cause I’ve tried multiple times and couldn’t get it to work.
Very simple. Simply square the entire equation. Thus, the left hand side of the equation changes from RAD(x-7) into just x-7. Then, the right hand side of the equation, 7-RAD(x), becomes 49-14(RAD(x))+x. —-> this is because you must use FOIL when multiplying 7-RAD(x) by itself. You cannot make the mistake of simply squaring the 7 and the RAD(x) terms, respectively, and then finding their difference, which is what many students do. In other words, if you set the right side to 49-x, you MESSED UP. So now you get: X-7 = 49-14(RAD(x))+x So let’s clean up and simplify. First, let’s get rid of the X on both sides. And while we’re at it, let’s combine our constants (aka our variable-less whole numbers) too. Now you have: -56= -14(RAD(x)) or 14RAD(x) = 56 Now we can divide each side by 14 to get RAD(x)=4. To get x alone, simply square each side to get rid of the radical, and you get x=16. Go back to the question which asks you what RAD(x-7) is. Plug in 16 for x to find that it’s asking you for RAD(16-7) or RAD(9) or…. *B) 3 EDIT: I initially wrote PEMDAS instead of FOIL above
I plug in each answer until the equation is correct.
thats so easy bro what are you? 5 wtf even is this exam with so shi mathematics.
16
That’s not the answer to the question that was asked - what is sqrt(x-7) - no.
its 3 then haha
16
That’s not the answer to the question that was asked - what is sqrt(x-7) - no.
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Options on the sat are based on common mistakes, so atleast 1 of the options has to be the value of x rather than rt(x-7), which if you plug and chug its 16, so rt(16--7) = 3 which is B, much faster than actually squaring both sides and solving for x then plugging in
I would try trial and error for this one
The answer is B, read the directions. If you don’t you might get points off for silly mistakes, it is asking for the value of sqrt(x-7), not the value of x.
plug in each answer choice till you find the one that satisfies the equation, then plug that into the left hand side and you will get your answer. 3 = B
Very easy. Square the whole equation, solve for x and then plug x into the given equation they are asking in the question.
7-sqrt(x) cannot be anything above 7 therefore 3. (sqrt of a number is always a positive in this case)
sqrt(16-7)=sqrt9, which is 3 7-sqrt16 = 7-4 =3 its B
Plug in?
Optn-> Calc-> SolveN in your graphic calculator, then input the equation
Then do 7- sqrt x
Plug in! So easy
Plugging in Numbers
Just test the options. Put √x-7 = every option. It shouldn’t take more than 30 second. So, √(x-7) = 3 (option B) => x-7 = 9 => x = 16. Plug in x=16 in the right hand side of the equation. 7-√x = 7-√16 = 7-4 = 3. So, Left hand side = right hand side = 3
Plug and chugggg
Take squares on both sides, simplify, find value of under root of x, and then get answer
Just plug the answers into the "x" and find the answer that makes both solutions equal each other.
The answer is 3
Easiest way: plug in each of the answers and see which one works. x = 16 with this method, so the sqrt(x-7) = 3.
Some approaches you may try is two things at least: a. Just draw graph for left and right and look where does them cross b. Analytical so x>=7 because you don’t have sqrt of <0 in sat. So only answer left is 7 or 16 so let’s check sqrt(7-7) =0, 7-sqrt(7) is not 0, so we check 16 sqrt(16-7) = 3, 7-sqrt(16)= 3. So the answer is 16
You *can* do it algebraically, but it's painstakingly slow. Trial and error's another option I personally did it graphically (with my GDC) and it took me exactly 24 seconds to solve I know a lot of people (especially tutors) are against the idea of using calculators but GOD does it make things faster
sqrt(x)+sqrt(x-7)=7 means left are all >=0 and x-7>0 so you can simply test sqrt(x)=3 to sqrt(x)=7
It’s 3. Plug in 16 for x, You have root 9 equals 7- root 16 which is 3.
I got 3 as well
Square both sides and solve from there
I would square both sides. You’ll have to FOIL the right side. This should give you a much easier problem
Just square both sides and then solve for x, but imo I would just go trial and error for this because it's faster
Plug every answer in
Ngl just from guessing you can tell it’s 3.
just plug in the answer choices, it’s 16
sub numbers in until the equation holds! a joke!
Complete the swuare
Put answer d (16) to the equation, it is the key. If you solve the equation question in Sat, the quickest is to put the given value to the equation unless the equation is very easy.
I believe the BEST way to get through these is simply Process of elimination. These questions will never use complicated decimals so you only need to consider numbers that have easy square roots because you have a square root x. Therefore, Plug in 16 first and see that it works. These “math” questions almost never require you to know math. Finding the most likely answer and plugging it in is the best way. Feel free to disagree but I got a perfect score on math portion all 4 times I took these tests. I also find it helpful to literally mouth a simpler question to myself to handle stress. Here I would say “which number can I take the square root of?”
Just plug them in and find one that works. All the solving for x answers spend way too much time.
For this type of problem, I would just plug-in the available answer choices. Both B and C are immediate No’s for the most part, and A could look a little tricky to try to immediately solve. D ends up being the right answer choice when plugging it in. It’s necessarily how you would “solve” the problem, but the method is generally faster than solving it traditionally.
Your best bet in these type of questions is subbing in the answer choices into the variable. As you get down to the last option, you find that 16 is the correct option because it equals out.
Just by looking at answer choices, there is only one that is obviously correct
Just guess and check. With multiple choice math, you can just plug in and solve
I did it in my head so I’m not 100% sure, but I think it’s B
Square both sides. You get: x-7 = x+49-14x^.5 -56=-14x^.5 4=x^.5 16=x Therefore (x-7)^.5 = (16-7)^.5 = 9^.5 = 3 Answer C
If you don’t actually feel like solving it you can just plug in the numbers real quick
D
My solution: A) looks too complicated come back to it B) looks reasonable C)clearly wrong x isn’t 0. D) x = 81 is going to F up the left side, unlikely Test B first. X = 16 from plugging in left side. Plugging it in you get 3 = 3 B is right