Question

Find the sum of the squares of the roots of 3x^2 + 4x + 12 = 0.

Answer 1

Step-by-step explanation:

Considering the equation

$3x^2\:+\:4x\:+\:12\:=\:0$

$\mathrm{Quadratic\:Equation\:Formula:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=3,\:b=4,\:c=12:\quad x_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:3\cdot \:12}}{2\cdot \:3}$

$x=\frac{-4+\sqrt{4^2-4\cdot \:3\cdot \:12}}{2\cdot \:3}$

   $=\frac{-4+\sqrt{128}i}{6}$

   $=\frac{-4+8\sqrt{2}i}{6}$

   $=\frac{4\left(-1+2i\sqrt{2}\right)}{6}$

   $=-\frac{2}{3}+\frac{4\sqrt{2}}{3}i$

similarly,

$x=\frac{-4-\sqrt{4^2-4\cdot \:3\cdot \:12}}{2\cdot \:3}:\quad -\frac{2}{3}-i\frac{4\sqrt{2}}{3}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=-\frac{2}{3}+i\frac{4\sqrt{2}}{3},\:x=-\frac{2}{3}-i\frac{4\sqrt{2}}{3}$

Finding the sum of the squares of the roots

$\left(-\frac{2}{3}+i\frac{4\sqrt{2}}{3}\right)^2+\left(-\frac{2}{3}-i\frac{4\sqrt{2}}{3}\right)^2$

as

$\left(-\frac{2}{3}+i\frac{4\sqrt{2}}{3}\right)^2=\frac{-28-16\sqrt{2}i}{3^2}$

and

$\left(-\frac{2}{3}-i\frac{4\sqrt{2}}{3}\right)^2=\frac{-28+16\sqrt{2}i}{3^2}$

so

$=\frac{-28-16\sqrt{2}i}{3^2}+\frac{-28+16\sqrt{2}i}{3^2}$

$\mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-28-16\sqrt{2}i-28+16\sqrt{2}i}{3^2}$

$=\frac{-56}{3^2}$             ∵ $-28-16\sqrt{2}i-28+16\sqrt{2}i=-56$

$=-\frac{56}{9}$

Answer 2

Answer:

$- \frac{56}{9}$

Step-by-step explanation:

The given quadratic equation is

$3 {x}^{2} + 4x + 12 = 0$

Where a=3, b=4, and c=12.

Let m and n be the roots.

The sum of the squares of the roots becomes:

${m}^{2} + {n}^{2} = {(m + n)}^{2} - 2mn$

This implies that:

${m}^{2} + {n}^{2} = {( - \frac{b}{a} )}^{2} - 2( \frac{c}{a} )$

Substitute a=3, b=4 and c=12 to get;

${m}^{2} + {n}^{2} = {( - \frac{4}{3} )}^{2} - 2( \frac{12}{3} )$

We evaluate to get:

${m}^{2} + {n}^{2} = \frac{16}{9} - 2( 4)$

${m}^{2} + {n}^{2} = - \frac{56}{9}$