(solved)Question 11 SSC-CGL 2020 March 4 Shift 3

Question 11
SSC-CGL 2020 Mar 4 Shift 3

If 12x² - 21x + 1 = 0, then the value of 9x² + 1/16x² =?

Solution in Detail

Divide both sides by $\displaystyle 4x$

$\displaystyle \frac{12x^2}{4x} - \frac{21x}{4x} + \frac{1}{4x} = 0$

$\displaystyle \therefore 3x - \frac {21}{4} + \frac{1}{4x} = 0$

$\displaystyle \therefore 3x + \frac{1}{4x} = \frac {21}{4}$

Squaring, we get

$\displaystyle (3x)^2 + \frac 32+ \bigg(\frac{1}{4x}\bigg)^2 = \bigg(\frac{21}{4}\bigg)^2$

$\displaystyle 9x^2 + \frac{1}{16x^2} = \frac{417}{16}\:\underline{Ans}$