Answer :
Answer:
[tex]\frac{-6+\sqrt{78} }{6}[/tex] and [tex]\frac{-6-\sqrt{78} }{6}[/tex]
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
[tex]\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}[/tex]
Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be
[tex]\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6}, \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}[/tex]
= [tex]\frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}[/tex]
=[tex]\frac{-6+\sqrt{78} }{6}[/tex] and [tex]\frac{-6-\sqrt{78} }{6}[/tex]
Hence, x= [tex]\frac{-6+\sqrt{78} }{6}[/tex]
and x= [tex]\frac{-6-\sqrt{78} }{6}[/tex]
(Answer)