Respuesta :
Answer:
The real roots are
[tex]x=\frac{(-3+\sqrt{37})}{4}[/tex] and [tex]x=\frac{(-3-\sqrt{37})}{4}[/tex]
The sum of the squares of these roots is [tex]\frac{-3}{2}[/tex]
Step-by-step explanation:
The given quadratic equation is [tex]8x^2+12x-14[/tex] has two real roots.
To find the roots .
[tex]8x^2+12x-14=0[/tex]
Dividing the above equation by 2
[tex]\frac{1}{2}(8x^2+12x-14)=\frac{0}{2}[/tex]
[tex]4x^2+6x-7=0[/tex]
For quadratic equation [tex]ax^2+bx+c=0[/tex] the solution is [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Where a and b are coefficents of [tex]x^2[/tex] and x respectively, c is a constant.
For given quadratic equation
a=4, b=6, c=-7
[tex]x=\frac{-6\pm\sqrt{6^2-4(4)(-7)}}{2(4)}[/tex]
[tex]=\frac{-6\pm\sqrt{36+112}}{8}[/tex]
[tex]=\frac{-6\pm\sqrt{148}}{8}[/tex]
[tex]=\frac{-6\pm\sqrt{37\times 4}}{8}[/tex]
[tex]=\frac{-6\pm\sqrt{37}\times\sqrt{4}}{8}[/tex]
[tex]=\frac{-6\pm\sqrt{37}\times 2}{8}[/tex]
[tex]=2\frac{(-3\pm\sqrt{37})}{8}[/tex]
[tex]=\frac{-3\pm\sqrt{37}}{4}[/tex]
[tex]x=\frac{(-3\pm\sqrt{37})}{4}[/tex]
The real roots are
[tex]x=\frac{(-3+\sqrt{37})}{4}[/tex] and [tex]x=\frac{(-3-\sqrt{37})}{4}[/tex]
Now to find the sum of the squares of these roots
[tex]\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3+\sqrt{37}-3-\sqrt{37}}{4}[/tex]
[tex]=\frac{-6}{4}[/tex]
[tex]=\frac{-3}{2}[/tex]
[tex]\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3}{2}[/tex]
Therefore the sum of the squares of these roots is [tex]\frac{-3}{2}[/tex]
