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

First recognize that \(x^2 – 10x + 25=(x-5)(x-5)=(x-5)^2\)
So, we can rewrite the given equation as follows: \((x^2 – 25)^2 = (x-5)^2\)

Do you see the difference of squares hiding in this equation?

Key property (difference of squares factoring): \(a^2 - b^2 = (a+b)(a-b)\)

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 \(x=-4\)

Answer: D

Cheers,
Brent
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Response:

As a matter of fact, the question can be solved that way too; however, square root of (x^2 - 25)^2 is not x^2 - 25 but it is |x^2 - 25|. Similarly for (x - 5)^2. If you use the correct expressions for the square roots, you’ll obtain:

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

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

x^2 - 25 = x - 5

or

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

Solving the first equation x^2 - 25 = x - 5, one obtains the solutions x = 5 and x = -4 just like you did.

Solving the second equation x^2 - 25 = -(x - 5), we get:

x^2 - 25 = -x + 5

x^2 + x - 30 = 0

(x + 6)(x - 5) = 0

x = -6 or x = 5

Combining, we obtain the solutions x = 5, x = -4 and x = -6. As you can see, the same solutions can also be obtained by taking square roots; however, you need to remember that the square root of y^2 is |y|, not y. Otherwise, you’ll be unable to obtain some of the solutions (like the way your solution failed to produce x = -6).
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\((x^2-25)^2 = x^2-10x+25\)

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

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

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

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

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

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

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

x= 5, -6 or -4

D

\((x^2 – 25)^2 = (x—5)^{2} \)

\((x^{2} —25 —x+5)( x^{2} —25 +x—5) = 0\)

\((x^{2} —x —20)( x^{2} +x —30) =0\)

\((x—5)(x+4)(x—5)(x+6) =0\)

—>\( x=5\), \(x= —4\) and \(x= —6\)

Answer (D)

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

\((x^2 – 25)^2 = x^2 – 10x + 25 = (x-5)^2\)
\((x-5)^2(x+5)^2 = (x-5)^2\)
\((x-5)^2\{(x+5)^2-1\}=0\)
\((x-5)^2(x+4)(x+6)=0\)
x = {-6,-4,5}; 3 real solutions

IMO D
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I have one question here, why couldn't we eliminate X-5)^2 from RHS and LHS

(x-5) could be equal to 0. Eliminating (x-5)^2 from both sides would mean that you are possibly dividing both sides by 0 and division by 0 is undefined

Posted from my mobile device
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Simplifying, we have:

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

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

Notice that (x - 5)^2 is a common factor of both terms on the left side of the equation.

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

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

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

x = 5 or x = -6 or x = -4

We see that the equation has 3 real solutions.

Answer: D
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hi experts,

why couldn't we do as follows:

(x^2 - 25)^2 = (x - 5)^2
taking sq root of both sides
x^2-25=x-5
x^2-x-20=0
(x-5)(x+4)=0
sq rooting both sides
x^2-25=x-5
solving the quad we get x=5 and -4

why is it wrong to take the sq root of both sides here? ScottTargetTestPrep BrentGMATPrepNow

Kritisood

If \((x^2 - 25)^2 = (x - 5)^2\), there are 2 possibilities.

1. \(x^2-25 = (x-5)\)
2. \((x^2-25) = -(x-5)\)

You've already solved the first one. Solve 2nd one and you'll get the missing root.

In 3rd row, there are multiplication signs between terms, how can we take one term on LHS with subtraction operation? shouldnt that be a division operation?
Please tell me what i am missing?

Regards,
Gajanan

I have highlighted the portion you're referring to.

It's always okay to subtract the same quantity from both sides of the equation.
For example, if 2x + 1 = 11
It's perfectly fine to subtract 1 from both sides to get: 2x = 10

Likewise, the product (x-5)(x-5) represents some quantity.
So it's perfectly acceptable to subtract that quantity from both sides of the equation.

Here's an example:
It's true to write: 24 = (3)(8)
Notice that we can subtract the product (3)(8) from both sides to get an equally true equation: 24 - (3)(8) = 0

Has that help?
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\((x^2 – 25)^2 = x^2 – 10x + 25\)
\((x+5)^2(x-5)^2 = (x-5)^2\)

The equation above is valid if \(x-5=0\), with the result that \(x=5\).

If \(x-5≠0\), we can safely divide both sides by \((x-5)^2\).
The following equation is yielded:
\((x+5)^2 = 1\)

The equation above is valid if \(x+5=1\), with the result that \(x=-4\), or if \(x+5=-1\), with the result that \(x=-6\).

Thus, there are 3 real options for x:
5, -4, -6


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\((x^2 – 25)^2 = x^2 – 10x + 25\)

Rewrite the left side using the difference of squares: \([(x – 5)(x + 5)]^2\)

Rewrite the right side as it's the square of a difference: \((x – 5)^2\)

Rewrite the left side again, as order in multiplication doesn't matter: \((x – 5)^2 * (x + 5)^2\)

So we have: \((x – 5)^2 * (x + 5)^2 = (x – 5)^2\)

Now, either \((x - 5) = 0\) so one possible solution is: \(x = 5\)

OR ELSE we can divide both sides by \((x - 5)^2\), in which case:
\((x + 5)^2 = 1\), which means that \(|x + 5| = 1\), or if we prefer to have an absolute value of a difference:

\(|x - (-5)| = 1\)

We can conclude that x is located 1 unit away from (-5), so x is either (-4) or (-6).

I'm counting a total of three possible solutions for x (5, -4, or -6), so the answer is D.
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BrentGMATPrepNow

Curious to know what you think. Isn't there a chance that solving such questions using equations can give us extraneous solutions? Should we plug in the roots to check the validity of the solutions?

On the GMAT, the only times we have the potential for extraneous roots is when the equation features either the absolute value or the square root of some variable expression.
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BrentGMATPrepNow



Your solution was very straightforward and I like it. However, don't you agree that if we take roots on both sides of the original equation we end up with absolute values on both sides?

(x^2-25)^2 = x^2 - 10x + 25 Taking root on both the sides.

|(x^2 - 25)| = |(x-5)| We can opt to solve this question this way as well. This is the reason I asked if there's a chance whether we can end up with an extraneous solution.

Ahh, I see.
Yes, if you're going to inject square roots and/or absolute values into the equation, then it would be wise to test for extraneous roots.
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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.