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The set of solutions for the equation (x^2 – 25)^2 = x^2 – 10x + 25 co

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BrentGMATPrepNow
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   13 Dec 2022, 18:31
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woohoo921 wrote:
BrentGMATPrepNow

Thank you for your helpful response. I did not quite follow this part:
"(x - 5)^2 * [(x + 5 + 1)(x + 5 - 1)] = 0
(x - 5)^2 * (x + 6)(x + 4) = 0"

I did this instead
(x-5)^2[(x+5)^2 - 1]=0
I then simplified what was in the brackets to be (x+5)(x+5)-1 --> x^2 + 10x + 24

So, my final answer was (x-5)(x-5)(x+12)(x-10) ---> x=5,-12,10

Why is this incorrect? Thank you again.


Here's my full solution (for comparison):
Take: \((x^2 – 25)^2 = (x-5)^2\)

Subtract \((x-5)^2\) from both sides of the equation to get: \((x^2 – 25)^2 - (x-5)^2 = 0\) Hey, that's a difference of squares!

Apply the above property to write the following equivalent equation: \([(x^2 – 25) + (x-5)][(x^2 – 25) - (x-5)] = 0\)

Simplify to get: \([x^2 + x – 30][x^2 – x - 20] = 0\)

Factor: \([(x+6)(x-5)][(x-5)(x+4)] = 0\)

This means \((x+6)=0\) or \((x-5)=0\) or \((x+4)=0\), which means there are THREE solutions: \(x=-6\), \(x=5\) and \(x=-4\)

I'm having a hard time figuring out how you went from (x-5)^2[(x+5)^2 - 1] = 0 to (x+5)(x+5)-1 --> x^2 + 10x + 24.
Please advise.
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   19 Feb 2023, 06:19
Hey guys,

\((x − 25)^2 = x^2 − 10x + 25\)
\((x + 5)^2*(x − 5)^2 = (x − 5)^2\)

Why can't we divide both sides by \((x − 5)^2\)?
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   19 Feb 2023, 08:53
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echo9000 wrote:
Hey guys,

\((x − 25)^2 = x^2 − 10x + 25\)
\((x + 5)^2*(x − 5)^2 = (x − 5)^2\)

Why can't we divide both sides by \((x − 5)^2\)?


You cannot reduce \((x + 5)^2*(x − 5)^2 = (x − 5)^2\) by \((x − 5)^2\) because \((x − 5)^2\) can be 0 and we cannot divide by 0. By doing so you loose a root, namely \((x − 5)^2=0\) --> x = 5.

Never reduce equation by a variable (or expression with a variable), if you are not certain that the variable (or expression with the variable) doesn't equal to zero. We cannot divide by zero.
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   07 Mar 2023, 05:42
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parkhydel wrote:
The set of solutions for the equation \((x^2 – 25)^2 = x^2 – 10x + 25\) contains how many real numbers?

A. 0
B. 1
C. 2
D. 3
E. 4

PS16731.02


\((x^2 – 25)^2 = x^2 – 10x + 25\)

\([(x+5)(x-5)]^2= (x-5)^2\)

\((x+5)^2(x-5)^2= (x-5)^2\)

One solution is x=5, which will make both sides equal to 0.

If x≠5, then (x-5) is nonzero, allowing us to safely divide both sides by \((x-5)^2\):
\(\frac{(x+5)^2(x-5)^2}{(x-5)^2}= \frac{(x-5)^2}{(x-5)^2}\)
\((x+5)^2 = 1\)
The resulting equation is valid if x+5=1 or x+5=-1, yielding two additional values for x and bringing the total number of solutions to 3.

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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   15 Apr 2023, 15:07
Hi ScottTargetTestPrep

Quote:
(x^2 - 25)^2 = (x - 5)^2

|x^2 - 25| = |x - 5|

x^2 - 25 = x - 5


The third line is true when x^2 - 25 >= 0, so when x^2 >=25. This means the range of values of x are x<=-5 and x>=5. However, -4 does not meet this range and yet is the solution to the equation. A little lost here.
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   20 Apr 2023, 18:21
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Vegita wrote:
Hi ScottTargetTestPrep

Quote:
(x^2 - 25)^2 = (x - 5)^2

|x^2 - 25| = |x - 5|

x^2 - 25 = x - 5


The third line is true when x^2 - 25 >= 0, so when x^2 >=25. This means the range of values of x are x<=-5 and x>=5. However, -4 does not meet this range and yet is the solution to the equation. A little lost here.


You are absolutely correct that |x^2 - 25| is equal to x^2 - 25 when x is less than or equal to -5 or greater than or equal to 5; however, |x^2 - 25| = |x - 5| holds when x^2 - 25 and x - 5 have the same sign. Notice that when x = -4, both x^2 - 25 and x - 5 are negative, so the equation becomes -(x^2 - 25) = -(x - 5), which is equivalent to x^2 - 25 = x - 5.
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   12 Jan 2024, 15:10
nick1816 wrote:
\((x^2-25)^2 = x^2-10x+25\)

\((x+5)^2(x-5)^2 = (x-5)^2\)

\((x+5)^2(x-5)^2 - (x-5)^2 = 0\)

\((x-5)^2 [(x+5)^2 -1] = 0\)

\((x-5)^2 [x^2+10x+25 -1] = 0\)

\((x-5)^2 [x^2+10x+24 ] = 0\)

\((x-5)^2 [x^2+4x+6x+24 ] = 0\)

\((x-5)^2 (x+6)(x+4) = 0\)

x= 5, -6 or -4

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parkhydel wrote:
The set of solutions for the equation \((x^2 – 25)^2 = x^2 – 10x + 25\) contains how many real numbers?

A. 0
B. 1
C. 2
D. 3
E. 4

PS16731.02


Why the (x+5)^2 goes into [] and not the (x-5)^2?
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink] New post   13 Jan 2024, 14:30
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Thib33600 wrote:
nick1816 wrote:
\((x^2-25)^2 = x^2-10x+25\)

\((x+5)^2(x-5)^2 = (x-5)^2\)

\((x+5)^2(x-5)^2 - (x-5)^2 = 0\)

\((x-5)^2 [(x+5)^2 -1] = 0\)

\((x-5)^2 [x^2+10x+25 -1] = 0\)

\((x-5)^2 [x^2+10x+24 ] = 0\)

\((x-5)^2 [x^2+4x+6x+24 ] = 0\)

\((x-5)^2 (x+6)(x+4) = 0\)

x= 5, -6 or -4

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parkhydel wrote:
The set of solutions for the equation \((x^2 – 25)^2 = x^2 – 10x + 25\) contains how many real numbers?

A. 0
B. 1
C. 2
D. 3
E. 4

PS16731.02


Why the (x+5)^2 goes into [] and not the (x-5)^2?


We factor out (x-5)^2 from \((x+5)^2(x-5)^2 - (x-5)^2 = 0\) and get \((x-5)^2((x+5)^2 - 1) = 0\).
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Re: The set of solutions for the equation (x^2 25)^2 = x^2 10x + 25 co [#permalink]
13 Jan 2024, 14:30
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