Question

Find the quadratic equation whose roots are double the roots of the quadratic equation 2x² - 3x + 6 = 0.

Answer 1

Let \(\alpha\) and \(\beta\) are roots of 2x² - 3x + 6 = 0

\(\therefore\) Sum of roots = \(\alpha\) + \(\beta\) = \(\frac{-b}{a}\) = \(\frac{-(-3)}{2}=\frac32 \)

& product of roots = \(\alpha\)\(\beta\) = \(\frac ca = \frac 62=3\)

\(\therefore\) 2\(\alpha\) + 2\(\beta\) = 2(\(\alpha\) + \(\beta\)) = 2 x \(\frac32\) = 3

& 2\(\alpha\) x 2\(\beta\) = 4 \(\alpha\)\(\beta\)  = 4 x 3 = 12

\(\therefore\) Quadratic equation whose roots are  2\(\alpha\) & 2\(\beta\) is 

x² - (sum of roots)x + products of roots = 0

⇒ x² - (2\(\alpha\) + 2\(\beta\))x + 2\(\alpha\) x 2\(\beta\)  = 0

⇒ x² - 3x + 12 = 0

Hence, required quadratic equation is x² - 3x + 12 = 0