Question

If $\alpha$ is a root of polynomial $4x^2+2x-1$ and $f\left(x\right)=4x^2-3x+1$, then $2\left(f\right(\alpha )+(\alpha )=$

• A. $0$
• B. $-1$
• C. $1$
• D. none of these

Answer 1

Answer: (C)

$1$

According to question,

$$\begin{array}{c}4x^2+2x-1=0\\ x=\frac{-2\pm \sqrt{4+16}}{8}=\frac{-2\pm 2\sqrt{5}}{8}=\frac{-1\pm \sqrt{5}}{4}\\ \therefore \alpha =\frac{\sqrt{5}-1}{4}\\ f\left(x\right)=4x^3-3x+1\\ f\left(\alpha \right)=4\times {\left(\frac{\sqrt{5}-1}{4}\right)}^3-3\left(\frac{\sqrt{5}-1}{4}\right)+1\\ \Rightarrow \left(\frac{{(\sqrt{5}-1)}^2}{16}\right)-\frac{3\sqrt{5}-3}{4}+1\\ \Rightarrow \frac{5\sqrt{5}-1-3\sqrt{5}(\sqrt{5}-1)-12\sqrt{5}+12+16}{16}\\ \Rightarrow \frac{5\sqrt{5}-1-15+3\sqrt{5}-12\sqrt{5}+12+16}{16}\\ \Rightarrow \frac{-4\sqrt{5}+12}{16}=\frac{3-\sqrt{5}}{4}\\ \therefore f\left(\alpha \right)=\frac{3-\sqrt{5}}{4}\\ Now,\\ 2\left(f\right(\alpha )+\alpha )=2\left(\frac{3-\sqrt{5}}{4}+\frac{\sqrt{5}-1}{4}\right)\\ \Rightarrow 2\left(\frac{3-\sqrt{5}+\sqrt{5}-1}{4}\right)\\ \Rightarrow 2\left(\frac{2}{4}\right)=1\\ \therefore 2\left(f\right(\alpha )+\alpha )=1\\ so,thatcorrectoptionisC.\end{array}$$