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(x^2 – 25)^2 = x^2 – 10x + 25
RHS,
we use (a-b)^2, ie, (x-5)^2
(x^2 – 25)^2 =(x-5)^2
taking square root on both side,
x^2-25=x-5
x^2-x-20=0
(x+4)(x-5)=0
x=5,-4
shouldn't ans be c....

why is this not right ?

(X-5)(X-5)(X+5)(X+5) = (X-5)(X-5)

dividing both sides by (x-5)(x-5)
(x+5)^2 = 1
x = -4 & -6

This piece

\((x^2 – 25)^2 \)

is equal to this

\((x^2 – 25)(x^2 – 25) \)

Which is equal

to this

\((x+5)(x-5)(x+5)(x-5)\)

Check your calculations above guys

Sir, is not \( X^2 = a^2\) give \(x = +_a ?\) . I used this way and got 4 real roots.

Why can't we take square root on both sides? That gives us only two roots.

Yes I did this and got only two roots i.e. 5 and -4. Why is this not correct?

By doing it, we will need to consider + & - sign as well.

Can you please show us the calculation of yours? So we can help.

rx10 Yes you are right. We have to consider the + and - sign as we are introducing the sq. root. We get three roots.

Here in this step, when we solve for each solution, we set \((x-5)^2 = 0\) and then take the square root on both sides. Doesn't this mean that we have to consider both the positive and negative signs after taking the square root?

I found the error.
THANKS!

Could you please explain?

This can also be solved by square rooting both sides.

Then we'll get (x^2 - 5^2) = +or-(x-5)
And upon solving further, we get x=5 and x=-4 when +(x-5).
And x=5 and x=-6 when -(x-5) or -x+5.

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I tried solving this sum .
And i ended up getting:

(x^2 – 25)^2 = x^2 – 10x + 25
(x^2 – 25)^2 = (x-5)^2
I took square root on both sides and then i got this:
(x-5)(x+5) = (x-5)
x + 5 = 1
x = -4

What is wrong in the method?

First, there's a problem when you took the square root of both sides by simply removing the power of 2 from both sides of the equation.
For example, notice that (-3)^2 = (3)^2
If we remove the power of 2 from both sides of the equation, we get -3 = 3, which is incorrect.

In general, if k^2 = j^2, then it could be the case that k = j, or k = -j

So, once we have (x^2 – 25)^2 = (x - 5)^2, we must consider two possibilities: (x^2 – 25) = (x - 5) and (x^2 – 25) = -(x - 5)

Another problem has to do with taking (x-5)(x+5) = (x-5) and dividing both sides of the equation by (x - 5) to get x + 5 = 1
By doing this you may have accidentally divided both sides of an equation by 0.
Notice that x = 5 is a solution to the equation (x-5)(x+5) = (x-5)
Also notice that, when x = 5, (x-5) = 0, so when you divided by (x-5) you are accidentally dividing by 0.
More importantly, this resulted in missing one of the solutions (x = 5)

simplify the equation to x^4-x^2+10x+550=0;
using Descarte's theorem the number of sign changes of coefficient of x in f(X) and f(-x) gives the number of real solution:-
for f(x) no. of sign changes =2
for f(-x) no. of sign changes =1.
total real soln=3.

Hi Brent GreenlightTestPrep, why can't we cancel out (x−5)(x−5)at this stage from both sides below?

Factor both sides to get: (x+5)(x−5)(x+5)(x−5)=(x−5)(x−5)