Question

What are the zeros of the quadratic function f(x) = 6x2 + 12x - 7?​

Answer 1

Answer:

$\frac{-6+\sqrt{78} }{6}$ and $\frac{-6-\sqrt{78} }{6}$

Step-by-step explanation:

We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.

So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)

Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.

This formula gives if, ax² +bx +c =0, the the two roots of the equation are

$\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be

$\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6}, \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}$

= $\frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}$

=$\frac{-6+\sqrt{78} }{6}$ and $\frac{-6-\sqrt{78} }{6}$

Hence, x= $\frac{-6+\sqrt{78} }{6}$

and x= $\frac{-6-\sqrt{78} }{6}$

(Answer)