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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.

7a) The roots of the equation 4x^2 - 7x - 1 = 0 are G and H. Evaluate G^2+ H^2B) Write-example-1
User Fredbaba
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1 Answer

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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

User Nulse
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