Question

Plss tell me how to obtain another value of x in my solution....as its ans is (x=0,x=a+b) ....one value is obtained how to obtain another plss help me

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Answer 1

Answer:

There are a lot of mistakes in this solution.

So if you do it correctly you will get the right solution.

Answer 2

Step-by-step explanation:

Actually, there are a couple of mistakes near the beginning (for example, a²+b² ≠ (a+b)²) so we haven't really even obtained the x=0 solution yet!  Let's start from scratch then...

Starting with

$\frac{\textstyle x-a}{\textstyle x-b}+\frac{\textstyle x-b}{\textstyle x-a}=\frac{\textstyle a}{\textstyle b}+\frac{\textstyle b}{\textstyle a}$

multiply throughout by ab(x-a)(x-b) to get rid of the fractions...

$ab\bigl((x-a)^2+(x-b)^2\bigr) = (a^2+b^2)(x-a)(x-b)$

Now expand and tidy up to write each side as a quadratic in x...

$ab\bigl((x^2-2ax+a^2)+(x^2-2bx+b^2)\bigr) = (a^2+b^2)\bigl(x^2-(a+b)x+ab\bigr)\\\\\Rightarrow ab\bigl(2x^2-2(a+b)x+(a^2+b^2)\bigr)=(a^2+b^2)\bigl(x^2-(a+b)x+ab\bigr)\\\\\Rightarrow 2abx^2-2ab(a+b)x+ab(a^2+b^2)=(a^2+b^2)x^2-(a+b)(a^2+b^2)x+ab(a^2+b^2)$

Notice the constant terms ab(a²+b²) on each side cancel.  Putting all the other terms together on one side and factorizing now gives...

$(a^2+b^2-2ab)x^2 - \bigl((a+b)(a^2+b^2)-2ab(a+b)\bigr)x = 0\\\\\Rightarrow (a-b)^2x^2 - (a+b)(a^2+b^2-2ab)x = 0\\\\\Rightarrow (a-b)^2x^2 - (a+b)(a-b)^2x = 0\\\\\Rightarrow (a-b)^2x\bigl(x-(a+b)\bigr) = 0$

From here we see the solutions are:

(i) If a = b then the equation is satisfied by all values of x (cool!)

(ii) If a ≠ b then either x = 0 or x = a+b.

Hope this helps!