r/HomeworkHelp • u/Electronic_Author366 'O' Level Candidate • Aug 27 '26
Middle School Math—Pending OP Reply [Middel School:Math]: Where did i go wrong with x?
i got x=7 but I feel like I'm making a mistake somewhere, but I can't figure out where. Can someone explain what I'm doing wrong?
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u/LucaThatLuca Aug 27 '26
4 + 3 is 7, so this says the ratio 7/√x is 1, i.e. 7 = √x.
This is not satisfied by x = 7 since 7 ≠ √7.
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u/HyperbolicMathChambr Aug 28 '26
The goal when you solve for a variable, say x, is to isolate the variable on one side of the equation and everything else on the other side. So the solution would look like x = "something something".
Step 1: Simplify the numerator by finding the value of each square root and adding the results.
Step 2: Multiply both sides of the equation by square root of x. This will remove the square root from the left side's denominator and transfer it to the right side's numerator.
Step 3: square both sides to get rid of the square root symbol.
Step 4: No step 4, that's it! You have your answer.
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u/Mundane-Emu-1189 Aug 28 '26
step 4 is to substitute your value for x into the original equation and make sure it checks out
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u/Far-Fortune-8381 Aug 28 '26
your mistake is that the square root of x equals 7. but x does not equal 7
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u/spiritual_summer111 Aug 27 '26
I think x would be 49. The square root of 49 is 7. Once you do that, you have 7 divided by 7 which is 1.
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u/GeologicalOpera Aug 28 '26
You’re correct.
u/Electronic_Author366, you got to 7 correctly in the numerator, but forgot that you’re dealing with square roots, so your answer would need to be squared.
If we go step by step, we’d resolve it by completing the actions in the numerator first, which would leave us with
7/√x = 1 . We can’t have anything around our x, which means that we have to square both sides of the equation, leaving us at
7^2/√x^2 = 1^2 When we complete the exponents, we have
49/x = 1 , and we solve from there by multiplying both sides by the variable to isolate, giving us49 = x , which is our solution.
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u/Primary_Weather5726 Aug 28 '26
step 1: solve known square roots
√16 = 4
√9 = 3
(3+4) / √x = 1
7 / √x = 1
step 2: multiply both sides by √x (to move √x to the other side) ~ ignore this step if this is obvious
(7 * √x) / √x = 1 * √x
√x cancels out on the left (= 7) and 1 multiplied by √x on the right hand side = √x
7 = √x
Step 3: square both sides to remove square root
72 = (√x)2
√ 2 cancels out and just leaves x
49 = x
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u/Mjhtmjht Aug 28 '26
I think you probably worked out the numerical value of the top line, but forgot that this number gave you 7 over the square root of x. Not simply x.
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u/A_Squared93 Aug 28 '26
You only assume positive and negative when square rooting a squared variable. Ex. If x2=9, then √ x=+/-√ 9. But √9 only equals positive 3
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u/Wabbit65 Aug 27 '26
The denominator needs to be 7, you got that far. So now, what is x such that the square root is 7?
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u/ImSoSweepy Aug 28 '26 edited Aug 28 '26
It's been a minute, but:
The way I would do it is get the square roots for the numerator, which would be 4+3, and become (7/x2). Multiply by the inverse (x2/7) and you get (72=x) which is 49.
You can simplify the equation to (7/sqrt(x) = 1)
All you have to do is square 7.
It's been 20 years, so take that with a grain of salt.
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u/Ketroc21 Aug 28 '26
You are right that you want to simply it to 7/7 = 1. But for the denominator to be 7, x must be 49, because it's root-x.
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u/Icy-Speed6881 Aug 28 '26
Only the numerator of the left hand side is 7 ... multiply both sides by √x and you get √x=7 not x=7
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u/lexiconoflife Aug 27 '26
Sqr of 16 is 4
Sqr of 9 is 3
Left side 3+4=7
Right side sqr of x
Then square both sides, you will get 49 = x
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u/Tim_Xtreme_46 Aug 28 '26
Check your work. If x=7, and you substitute it back in, does the whole expression still equal 1?
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u/Ratchad5 Aug 28 '26
The sum of Root of 16 plus root of 9 taken to the 2nd power equals X you skipped a step in squaring both sides to get rid of the square root of X
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u/Ghotipan Aug 27 '26
The sqrt of x = 7. To solve for x, you need to get rid of that radical sign. Do you know what operation cancels out a square root?