Question

What are the zeros of the quadratic function f(x) = 6x2 + 12x – 7? x = –1 – StartRoot StartFraction 13 Over 6 EndFraction EndRoot and x = –1 + StartRoot StartFraction 13 Over 6 EndFraction EndRoot x = –1 – StartFraction 2 Over StartRoot 3 EndRoot EndFraction and x = –1 + StartFraction 2 Over StartRoot 3 EndRoot EndFraction x = –1 – StartRoot StartFraction 7 Over 6 EndFraction EndRoot and x = –1 + StartRoot StartFraction 7 Over 6 EndFraction EndRoot x = –1 – StartFraction 1 Over StartRoot 6 EndRoot EndFraction and x = –1 + StartFraction 1 Over StartRoot 6 EndRoot EndFraction

Answer 1

Answer:

The answer is b

Step-by-step explanation:

Answer 2

Answer:

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

Step-by-step explanation:

In this problem, the function is

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

The zeros of the function are the values of x for which the function is zero, so:

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

We start by dividing each term by 6:

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

Then we add +1 and -1 to the equation:

$x^2+2x+1-1-\frac{7}{6}=0$

The first 3 terms $x^2+2x+1$ are the square of a binomial, $(x+1)$, so this becomes

$(x+1)^2-1-\frac{7}{6}=0$

Which is equivalent to

$(x+1)^2-\frac{13}{6}=0$ (1)

Now we can use the following rule:

$x^2-a^2 = (x+a)(x-a)$

To rewrite (1) as:

$(x+1)^2-\frac{13}{6}=\\=(x+1+\sqrt{\frac{13}{6}})(x+1-\sqrt{\frac{13}{6}})=0$

This expression is equal to zero if either one of the two terms in the brackets is equal to zero. Therefore:

1)

$x+1+\sqrt{\frac{13}{6}}=0 \rightarrow x=-1-\sqrt{\frac{13}{6}}$

2)

$x+1-\sqrt{\frac{13}{6}}=0 \rightarrow x=-1+\sqrt{\frac{13}{6}}$