Answer:
[tex]\textsf{a)}\quad y=8x^2+2x-3[/tex]
[tex]\textsf{b)}\quad y=x^2-2x-4[/tex]
Step-by-step explanation:
Given information:
[tex]x^2+bx+c=0[/tex]
[tex]x_1+x_2=-b[/tex]
[tex]x_1x_2=c[/tex]
Part (a)
[tex]\textsf{If}\quad x_1=\dfrac12\quad\textsf{and}\quad x_2=-\dfrac34[/tex]
[tex]\begin{aligned}\implies -b & =\dfrac12+\left(-\dfrac34\right)\\ & = \dfrac24-\dfrac34\\ & =-\dfrac14\end{aligned}[/tex]
[tex]\begin{aligned}\implies c & =\dfrac12 \cdot \left(-\dfrac34\right)\\ & = \dfrac{1 (-3)}{2 \cdot 4}\\ & =-\dfrac38\end{aligned}[/tex]
Substituting the found values of b and c into [tex]x^2+bx+c=0[/tex]
[tex]\implies x^2+\dfrac14x-\dfrac38=0[/tex]
Multiply both sides by 8 so that coefficients are integers:
[tex]\implies 8x^2+2x-3=0[/tex]
Therefore, the final quadratic equation is:
[tex]y=8x^2+2x-3[/tex]
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Part (b)
[tex]\textsf{If}\quad x_1=1+\sqrt{5}\quad\textsf{and}\quad x_2=1-\sqrt{5}[/tex]
[tex]\begin{aligned}\implies -b & =(1+\sqrt{5})+(1-\sqrt{5})\\ & = 1+1+\sqrt{5}-\sqrt{5}\\ & =2\end{aligned}[/tex]
[tex]\begin{aligned}\implies c & =(1+\sqrt{5})(1-\sqrt{5})\\ & = 1-\sqrt{5}+\sqrt{5}-5\\ & =-4\end{aligned}[/tex]
Substituting the found values of b and c into [tex]x^2+bx+c=0[/tex]
[tex]\implies x^2-2x-4=0[/tex]
Therefore, the final quadratic equation is:
[tex]y=x^2-2x-4[/tex]
