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