Answer:
[tex]x=-1+\sqrt{\frac{13}{6}}\\x=-1-\sqrt{\frac{13}{6}}[/tex]
Step-by-step explanation:
In this problem, the function is
[tex]f(x)=6x^2+12x-7[/tex]
The zeros of the function are the values of x for which the function is zero, so:
[tex]6x^2+12x-7=0[/tex]
We start by dividing each term by 6:
[tex]x^2+2x-\frac{7}{6}=0[/tex]
Then we add +1 and -1 to the equation:
[tex]x^2+2x+1-1-\frac{7}{6}=0[/tex]
The first 3 terms [tex]x^2+2x+1[/tex] are the square of a binomial, [tex](x+1)[/tex], so this becomes
[tex](x+1)^2-1-\frac{7}{6}=0[/tex]
Which is equivalent to
[tex](x+1)^2-\frac{13}{6}=0[/tex] (1)
Now we can use the following rule:
[tex]x^2-a^2 = (x+a)(x-a)[/tex]
To rewrite (1) as:
[tex](x+1)^2-\frac{13}{6}=\\=(x+1+\sqrt{\frac{13}{6}})(x+1-\sqrt{\frac{13}{6}})=0[/tex]
This expression is equal to zero if either one of the two terms in the brackets is equal to zero. Therefore:
1)
[tex]x+1+\sqrt{\frac{13}{6}}=0 \rightarrow x=-1-\sqrt{\frac{13}{6}}[/tex]
2)
[tex]x+1-\sqrt{\frac{13}{6}}=0 \rightarrow x=-1+\sqrt{\frac{13}{6}}[/tex]
