Question

If $\alpha$ and $\beta$ are the roots of the equation $4x^2-5x+2=0$, find the equation whose roots are

$\alpha +\frac{1}{\alpha }$ and $\beta +\frac{1}{\beta }$.

• A. $8x^2+30x+29=0$
• B. $x^2-30x+29=0$
• C. $8x^2-30x+29=0$
• D. $x^2+30x+29=0$

Answer 1

The correct option is C $8x^2-30x+29=0$

The equation is: $4x^2-5x+2=0$

Sum of the roots = $\frac{5}{4}$

Product of the roots = $\frac{2}{4}=\frac{1}{2}$
If the roots are $\alpha +\frac{1}{\alpha },\beta +\frac{1}{\beta }$

Sum of roots = $\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }$

= $\alpha +\beta +\frac{\alpha +\beta }{\alpha \beta }$

= $\frac{5}{4}+\frac{\frac{5}{4}}{\frac{1}{2}}$

= $\frac{5}{4}+\frac{5}{2}$

$\frac{15}{4}$
Product of roots = $(\alpha +\frac{1}{\alpha })(\beta +\frac{1}{\beta })$

= $\alpha \beta +\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }$

= $\frac{1}{2}+\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }+2$

= $\frac{1}{2}+\frac{\frac{25}{16}-1}{\frac{1}{2}}+2$

= $\frac{1}{2}+\frac{9}{8}+2$

= $\frac{4+9+16}{8}$

= $\frac{29}{8}$
Hence.the equation in the standard form, $x^2-Sx+P=0$ can be written as:

=$x^2-\frac{15}{4}x+\frac{29}{8}=0$

= $8x^2-30x+29=0$

Hence option ${}^{\prime }{C}^{\prime }$ is the answer.