How do i put this is terms of "x"?

Question

(ax+b)^2 - (bx+a)^2 = 0 ; a does not equal =/-b

I broke down this problem, and winded up with 1=x^2 + x /x^2 + x .

The answer is x=+/- 1.

Did I do it correctly?

Answer 1

(ax+b)²-(bx+a)² = 0
a²x² + 2abx + b² - (b²x² + 2abx + a²) = 0
(a²-b²)x² + b²-a² = 0

Since a≠±b, a²≠b², so we can divide by a²-b²:

x² - 1 = 0
x=±1

So your answer is correct. I'm not sure how you ended up with 1=x² + 1/x + x as an intermediate step though, nor would you be able to obtain x=±1 as a result from that, so I suspect there is still an error in your derivation. You should endeavor to find it, since it is better to get the correct answer through a correct method than through luck.

Answer 2

Not sure I understand what you mean by put this in terms of x. If you are saying solve the above equation for x, then it is a long one which I will shorten down.

(ax)^2- (bx)^2 +b^2 -a^2 =0
(ax)^2- (bx)^2 = a^2 -b^2
x^2(a^2 -b^2)=a^2 -b^2-------divide both sides by a^2 -b^2
x^2=1
x=+/- sqrt 1
x= +/- 1



If you need further clarification and step by step, you can email me.

Answer 3

(ax+b)^2 - (bx+a)^2 = 0
((ax)^2 + 2abx +b2)-((bx)^2 + 2abx +a^2)=0
(ax)^2 - (bx)^2 +b^2 - a^2 = 0
x^2[a^2 - b^2] = a^2 - b^2
x^2 = [a^2 - b^2]/[a^2 - b^2] = 1
x^2 = 1
x = +/-1
Yes!!! you did it correctly.

Answer 4

How's about just by inspection.
You can lose a lot of time on a test
churning algebra.

Answer 5

if x=-1

we have

(-a+b)^2-(-b+a)^2=0

=(-a+b)^2+(-a+b)^2

=2(b-a)^2=0

no pasible

Answer 6

the answer is

x = -a/b