How to solve (x-2)^2=25

First of all you have to get rid of the exponent.
Any idea how ?
 
Hi,
Well, it looked like a quadratic equation to me, and I wrote out x^2-4 = 25
Then (x+2) (x-2)
x+2=0, x-2=0
x=-2, x=2. But I don't know what to do with the 25.

Or maybe x-5=0, x+5=0 because 5X5 equals 25.

I think it should be factored out and then set to 0.

That's what the other problems are like that I have been doing.

But when I see that exponent outside the parentheses it really throws me off.
 
Sorry, but this x^2-4 = 25 is wrong.
It must be (x-2)(x-2)=25.
On the other hand you can just use the square root.
 
Okay,

Yes, the problem says "Solve" (x-2)2=25

That might look clearer with the exponent mark above
 
Last edited:
Yes,
(x-2)2= (x-2)(x-2) = x2 - 4x + 4
or √(x-2)2= √25
That would be the first step. But you can easily see
what x must be with this equation: (x-2)(x-2) = 25
 
What serds (not "surds"?) is trying to get you to see is that 25=52\displaystyle 25= 5^2 so your equation is (x−2)2=52\displaystyle (x- 2)^2= 5^2. Go ahead and take the square root of both sides now.
 
(x-2)2 = 52

(x-2)2 - 52 = 0

now use:

a2 - b2 = (a+b)(a-b) → in your case (x-2) = a & 5 = b

so we get:

(x-2)2 - 52 = [(x-2) + (5)] * [(x-2) - (5)] = [x + 3] * [x - 7]

so, now you have

(x-2)2 - 52 = 0 → [x + 3] * [x - 7] = 0 ..... and continue....

I prefer this route - because it avoids confusion of introducing ± signs due to √ operation.
 
Steph Annie said:
Hi,
Well, it looked like a quadratic equation to me, and I wrote out x^2-4 = 25
Then (x+2) (x-2)
x+2=0, x-2=0
x=-2, x=2. But I don't know what to do with the 25.

Or maybe x-5=0, x+5=0 because 5X5 equals 25.

I think it should be factored out and then set to 0.

That's what the other problems are like that I have been doing.

But when I see that exponent outside the parentheses it really throws me off.
Click to expand...
You are correct that this is a quadratic equation. Good.

You have received excellent advice on very elegant ways to solve this problem, but it may be helpful to show some basic errors in your thinking.

(x−2)2 does not equal (x+2)(x−2). (x−2)2=x2−4x+4. And (x+2)(x−2)=x2−22=x2−4.\displaystyle (x - 2)^2\ does\ not\ equal\ (x + 2)(x - 2).\ (x - 2)^2 = x^2 - 4x + 4.\ And\ (x + 2)(x - 2) = x^2 - 2^2 = x^2 - 4.

I am sure you know that, but you are seeing lots of factoring in solving quadratics and forgot some basics.

The non-elegant way to solve this is to translate the quadratic equation into standard form.

Start by doing the indicated operation on the expression in parentheses.

(x−2)2=25  ⟹  x2−4x+4=25  ⟹  x2−4x+4−25=25−25  ⟹  x2−4x−21=0.\displaystyle (x - 2)^2 = 25 \implies x^2 - 4x + 4 = 25 \implies x^2 - 4x + 4 - 25 = 25 - 25 \implies x^2 - 4x - 21 = 0.

Now you can use any of the techniques that you have been studying, whether completing the square, factoring, or the quadratic formula. All of them work once you have put the quadratic into standard form.

None of them work until the equation is in standard form.
 
Thank you

I know it's been months since I first posted this, but I always meant to thank you for your answer.
It was very helpful and I appreciate it! Reading your explanation made things clearer, and felt like a math tutor was right there helping me.

Also, thank you to the others who replied as well with suggestions on how to solve the problem. It was all great! THANK YOU!



JeffM said:
You are correct that this is a quadratic equation. Good.

You have received excellent advice on very elegant ways to solve this problem, but it may be helpful to show some basic errors in your thinking.

(x−2)2 does not equal (x+2)(x−2). (x−2)2=x2−4x+4. And (x+2)(x−2)=x2−22=x2−4.\displaystyle (x - 2)^2\ does\ not\ equal\ (x + 2)(x - 2).\ (x - 2)^2 = x^2 - 4x + 4.\ And\ (x + 2)(x - 2) = x^2 - 2^2 = x^2 - 4.

I am sure you know that, but you are seeing lots of factoring in solving quadratics and forgot some basics.

The non-elegant way to solve this is to translate the quadratic equation into standard form.

Start by doing the indicated operation on the expression in parentheses.

(x−2)2=25  ⟹  x2−4x+4=25  ⟹  x2−4x+4−25=25−25  ⟹  x2−4x−21=0.\displaystyle (x - 2)^2 = 25 \implies x^2 - 4x + 4 = 25 \implies x^2 - 4x + 4 - 25 = 25 - 25 \implies x^2 - 4x - 21 = 0.

Now you can use any of the techniques that you have been studying, whether completing the square, factoring, or the quadratic formula. All of them work once you have put the quadratic into standard form.

None of them work until the equation is in standard form.
Click to expand...
 
This is a quadratic equation to solve

removing brackets: x^2 - 4x + 4 = 25
x^2 -4x -21 = 0 (quadratic form)

Factorizing: (x - 7)(x + 3) = 0
Answer: x = 7 or -3
 
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