Question

Please solve with explanation

Answer 1

Answer:

$\textsf{a)}\quad y=8x^2+2x-3$

$\textsf{b)}\quad y=x^2-2x-4$

Step-by-step explanation:

Given information:

$x^2+bx+c=0$

$x_1+x_2=-b$

$x_1x_2=c$

Part (a)

$\textsf{If}\quad x_1=\dfrac12\quad\textsf{and}\quad x_2=-\dfrac34$

$\begin{aligned}\implies -b & =\dfrac12+\left(-\dfrac34\right)\\ & = \dfrac24-\dfrac34\\ & =-\dfrac14\end{aligned}$

$\begin{aligned}\implies c & =\dfrac12 \cdot \left(-\dfrac34\right)\\ & = \dfrac{1 (-3)}{2 \cdot 4}\\ & =-\dfrac38\end{aligned}$

Substituting the found values of b and c into $x^2+bx+c=0$

$\implies x^2+\dfrac14x-\dfrac38=0$

Multiply both sides by 8 so that coefficients are integers:

$\implies 8x^2+2x-3=0$

Therefore, the final quadratic equation is:

$y=8x^2+2x-3$

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Part (b)

$\textsf{If}\quad x_1=1+\sqrt{5}\quad\textsf{and}\quad x_2=1-\sqrt{5}$

$\begin{aligned}\implies -b & =(1+\sqrt{5})+(1-\sqrt{5})\\ & = 1+1+\sqrt{5}-\sqrt{5}\\ & =2\end{aligned}$

$\begin{aligned}\implies c & =(1+\sqrt{5})(1-\sqrt{5})\\ & = 1-\sqrt{5}+\sqrt{5}-5\\ & =-4\end{aligned}$

Substituting the found values of b and c into $x^2+bx+c=0$

$\implies x^2-2x-4=0$

Therefore, the final quadratic equation is:

$y=x^2-2x-4$