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natulia [17]
3 years ago
13

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

Mathematics
2 answers:
Neko [114]3 years ago
6 0

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}

<u>Finding the sum of the squares of the roots</u>

\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}

Gnoma [55]3 years ago
4 0

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}

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