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

x = –1 – and x = –1 +
x = –1 – and x = –1 +
x = –1 – and x = –1 +
x = –1 – and x = –1 +

2 Answers

5 votes

Answer:


x=-1+(√(78))/(6),
x=-1-(√(78))/(6).

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=(-12\pm √(144-4* 6* (-7)))/(2* 6)


=(-12\pm √(144+168))/(12)=(-12\pm √(312))/(12)
=(-12\pm √(4* 78))/(12)=-1\pm (2√(78))/(12)


=-1\pm (√(78))/(6)


x=-1+(√(78))/(6)

and
x=-1-(√(78))/(6)

User Weisj
by
8.2k points
3 votes

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{(13)/(6)}


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


x1=-1+\sqrt{(13)/(6)}


x2=-1-\sqrt{(13)/(6)}

therefore

the answer is

The zeros of the quadratic function are


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


User Solomon Duskis
by
8.6k points

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