ANSWER:
x-intercepts of [tex]\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})[/tex]
SOLUTION:
Given, [tex]f(x)=x^{2}+12 x+24[/tex] -- eqn 1
x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.
Now, let us find the zeroes using quadratic formula for f(x) = 0.
[tex]X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Here, for (1) a = 1, b= 12 and c = 24
[tex]X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}[/tex]
[tex]\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}[/tex]
Hence the x-intercepts of [tex]\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})[/tex]
