Answer:
[tex] - \frac{56}{9} [/tex]
Step-by-step explanation:
The given quadratic equation is
[tex]3 {x}^{2} + 4x + 12 = 0[/tex]
Where a=3, b=4, and c=12.
Let m and n be the roots.
The sum of the squares of the roots becomes:
[tex] {m}^{2} + {n}^{2} = {(m + n)}^{2} - 2mn [/tex]
This implies that:
[tex] {m}^{2} + {n}^{2} = {( - \frac{b}{a} )}^{2} - 2( \frac{c}{a} )[/tex]
Substitute a=3, b=4 and c=12 to get;
[tex]{m}^{2} + {n}^{2} = {( - \frac{4}{3} )}^{2} - 2( \frac{12}{3} )[/tex]
We evaluate to get:
[tex]{m}^{2} + {n}^{2} = \frac{16}{9} - 2( 4)[/tex]
[tex]{m}^{2} + {n}^{2} = - \frac{56}{9} [/tex]
Step-by-step explanation:
Considering the equation
[tex]3x^2\:+\:4x\:+\:12\:=\:0[/tex]
[tex]\mathrm{Quadratic\:Equation\:Formula:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\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}[/tex]
[tex]x=\frac{-4+\sqrt{4^2-4\cdot \:3\cdot \:12}}{2\cdot \:3}[/tex]
[tex]=\frac{-4+\sqrt{128}i}{6}[/tex]
[tex]=\frac{-4+8\sqrt{2}i}{6}[/tex]
[tex]=\frac{4\left(-1+2i\sqrt{2}\right)}{6}[/tex]
[tex]=-\frac{2}{3}+\frac{4\sqrt{2}}{3}i[/tex]
similarly,
[tex]x=\frac{-4-\sqrt{4^2-4\cdot \:3\cdot \:12}}{2\cdot \:3}:\quad -\frac{2}{3}-i\frac{4\sqrt{2}}{3}[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=-\frac{2}{3}+i\frac{4\sqrt{2}}{3},\:x=-\frac{2}{3}-i\frac{4\sqrt{2}}{3}[/tex]
Finding the sum of the squares of the roots
[tex]\left(-\frac{2}{3}+i\frac{4\sqrt{2}}{3}\right)^2+\left(-\frac{2}{3}-i\frac{4\sqrt{2}}{3}\right)^2[/tex]
as
[tex]\left(-\frac{2}{3}+i\frac{4\sqrt{2}}{3}\right)^2=\frac{-28-16\sqrt{2}i}{3^2}[/tex]
and
[tex]\left(-\frac{2}{3}-i\frac{4\sqrt{2}}{3}\right)^2=\frac{-28+16\sqrt{2}i}{3^2}[/tex]
so
[tex]=\frac{-28-16\sqrt{2}i}{3^2}+\frac{-28+16\sqrt{2}i}{3^2}[/tex]
[tex]\mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}[/tex]
[tex]=\frac{-28-16\sqrt{2}i-28+16\sqrt{2}i}{3^2}[/tex]
[tex]=\frac{-56}{3^2}[/tex] ∵ [tex]-28-16\sqrt{2}i-28+16\sqrt{2}i=-56[/tex]
[tex]=-\frac{56}{9}[/tex]
