Question

solve 9x2-6ax+(a2-b2)=0

Answer 1

Hii dear here is your answer hope it helps you!!

:-)

Answer 2

Answer: The value of x is given below:

$\frac{1}{3}(1\pm \sqrt{1-a^2+b^2})$

Step-by-step explanation:

Since we have given that

$9x^2-6ax+(a^2-b^2)=0$

We need to find the roots :

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\frac{6\pm \sqrt{36-4\times 9\times (a^2-b^2)}}{2\times 9}\\\\x=\frac{6\pm \sqrt{36(1-a^2+b^2})}{18}\\\\x=\frac{6\pm 6\sqrt{1-a^2+b^2}}{18}\\\\x=\frac{1}{3}(1\pm \sqrt{1-a^2+b^2})$

Hence, the value of x is given below:

$\frac{1}{3}(1\pm \sqrt{1-a^2+b^2})$