Question

7a) The roots of the equation 4x^2 - 7x - 1 = 0 are G and H. Evaluate G^2+ H^2B) Write the equation of a quadratic with integer coefficients whose solutions are G^2 and H^2.Pls see the pic for more detail.

Answer 1

Given:

`4x^2-7x-1=0`

Solve:

Quadratic formula:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where,

`ax^2+bx+c=0`

Compaire the equation then:

`\begin{gathered} ax^2+bx+c=0 \\ 4x^2-7x-1=0 \\ a=4,b=-7,c=-1 \end{gathered}`

So roots of equation is:

`\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4(4)(-1)}}{2(4)} \\ x=\frac{7\pm\sqrt[]{49+16}}{8} \\ x=\frac{7\pm\sqrt[]{65}}{8} \end{gathered}`

So value of G and H is:

`\begin{gathered} G=\frac{7+\sqrt[]{65}}{8};H=\frac{7-\sqrt[]{65}}{8} \\ G=(7)/(8)+\frac{\sqrt[]{65}}{8};H=(7)/(8)-\frac{\sqrt[]{65}}{8} \end{gathered}`

So:

`\begin{gathered} =G^2+H^2 \\ =((7)/(8)+\frac{\sqrt[]{65}}{8})^2+((7)/(8)-\frac{\sqrt[]{65}}{8})^2 \\ =((7)/(8))^2+(\frac{\sqrt[]{65}}{8})^2+2((7)/(8))(\frac{\sqrt[]{65}}{8})+((7)/(8))^2+(\frac{\sqrt[]{65}}{8})^2-2((7)/(8))(\frac{\sqrt[]{65}}{8}) \\ =2((49)/(64)+(65)/(64)) \\ =2((114)/(64)) \\ =(114)/(32) \\ =3.5625 \end{gathered}`

(B)

If roots is a and b the equation is:

`x^2-(a+b)x+ab=0`

Then equation is:

`G^2+H^2=3.5625`

`\begin{gathered} G^2H^2=((7)/(8)+\frac{\sqrt[]{65}}{8})^2((7)/(8)-\frac{\sqrt[]{65}}{8})^2 \\ =(0.875+1.00778)^2(0.875-1.00778)^2 \\ =(3.54486)(0.01763) \\ =0.0624 \end{gathered}`

So equation is:

`x^2-3.5625x+0.0624`