Question

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

Answer 1

Answer:

$x=-1+\frac{\sqrt{78}}{6}$, $x=-1-\frac{\sqrt{78}}{6}$.

Step-by-step explanation:

The given quadratic equation function is f(x) = 6x²+12x -7

To find the zeros of the quadratic equation we rewrite the function as

6x² + 12x - 7 = 0

Now we know the quadratic formula to get the solutions as

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

here a = 6

b = 12

c = -7

By putting these values in the formula

$x=\frac{-12\pm \sqrt{144-4\times 6\times (-7)}}{2\times 6}$

$=\frac{-12\pm \sqrt{144+168}}{12}=\frac{-12\pm \sqrt{312}}{12}$$=\frac{-12\pm \sqrt{4\times 78}}{12}=-1\pm \frac{2\sqrt{78}}{12}$

$=-1\pm \frac{\sqrt{78}}{6}$

$x=-1+\frac{\sqrt{78}}{6}$

and $x=-1-\frac{\sqrt{78}}{6}$

Answer 2

we have

$f(x) = 6x^{2} + 12x -7$

To find the zeros equate the function to zero

$6x^{2} + 12x -7=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$7= 6x^{2} + 12x$

Factor the leading coefficient

$7= 6(x^{2} + 2x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$7+6= 6(x^{2} + 2x+1)$

$13= 6(x^{2} + 2x+1)$

Rewrite as perfect squares

$13= 6(x+1)^{2}$

$(13/6)=(x+1)^{2}$

square root both sides

$x+1=(+/-)\sqrt{\frac{13}{6}}$

$x=-1(+/-)\sqrt{\frac{13}{6}}$

$x1=-1+\sqrt{\frac{13}{6}}$

$x2=-1-\sqrt{\frac{13}{6}}$

therefore

the answer is

The zeros of the quadratic function are

$x=-1+\sqrt{\frac{13}{6}}$ and $x=-1-\sqrt{\frac{13}{6}}$