Answer:
option (b) is correct.
\(x_1=\frac{-1+3\sqrt{2}}{3},\:x_2=-\frac{1+3\sqrt{2}}{3}\)
Step-by-step explanation:
Given equation \(9x^2+6x-17=0\)
We have to solve the given quadratic equation using quadratic formula.
For the given standard quadratic equation \(ax^2+bx+c=0\)
Quadratic formula is given as \(x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
Comparing with given quadratic equation, we have a = 9, b = 6 and c = -17
Substitute in quadratic formula, we have,
\(x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:9\left(-17\right)}}{2\cdot \:9}\)
Simplify , we have \(\sqrt{6^2+4\cdot \:9\cdot \:17}=\sqrt{648}\)
we get,
\(x_{1,\:2}=\frac{-6\pm\sqrt{648}}{2\cdot \:9}\)
Also, \(\sqrt{648} =\sqrt{3^4\cdot \:2^3}=18\sqrt{2}\)
we have,
\(x_{1,\:2}=\frac{-6\pm 18\sqrt{2}}{2\cdot \:9}\)
Thus, seperating both facotors, we have,
\(x_{1}=\frac{-6\+18\sqrt{2}}{2\cdot \:9}\) and \(x_{2}=\frac{-6\-18\sqrt{2}}{2\cdot \:9}\)
Simplify, both we get,
\(x_1=\frac{-1+3\sqrt{2}}{3},\:x_2=-\frac{1+3\sqrt{2}}{3}\)
Thus, option (b) is correct.