we have
[tex]f(x) = 6x^{2} + 12x -7[/tex]
To find the zeros equate the function to zero
[tex] 6x^{2} + 12x -7=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]7= 6x^{2} + 12x[/tex]
Factor the leading coefficient
[tex]7= 6(x^{2} + 2x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]7+6= 6(x^{2} + 2x+1)[/tex]
[tex]13= 6(x^{2} + 2x+1)[/tex]
Rewrite as perfect squares
[tex]13= 6(x+1)^{2}[/tex]
[tex](13/6)=(x+1)^{2}[/tex]
square root both sides
[tex]x+1=(+/-)\sqrt{\frac{13}{6}}[/tex]
[tex]x=-1(+/-)\sqrt{\frac{13}{6}}[/tex]
[tex]x1=-1+\sqrt{\frac{13}{6}}[/tex]
[tex]x2=-1-\sqrt{\frac{13}{6}}[/tex]
therefore
the answer is
The zeros of the quadratic function are
[tex]x=-1+\sqrt{\frac{13}{6}}[/tex] and [tex]x=-1-\sqrt{\frac{13}{6}}[/tex]
Answer:
[tex]x=-1+\frac{\sqrt{78}}{6}[/tex], [tex]x=-1-\frac{\sqrt{78}}{6}[/tex].
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
[tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]
here a = 6
b = 12
c = -7
By putting these values in the formula
[tex]x=\frac{-12\pm \sqrt{144-4\times 6\times (-7)}}{2\times 6}[/tex]
[tex]=\frac{-12\pm \sqrt{144+168}}{12}=\frac{-12\pm \sqrt{312}}{12}[/tex][tex]=\frac{-12\pm \sqrt{4\times 78}}{12}=-1\pm \frac{2\sqrt{78}}{12}[/tex]
[tex]=-1\pm \frac{\sqrt{78}}{6}[/tex]
[tex]x=-1+\frac{\sqrt{78}}{6}[/tex]
and [tex]x=-1-\frac{\sqrt{78}}{6}[/tex]
