Question

The solution of the cubic $24x^3-14x^2-63x+45=0$. one root being double the other is

• A. $\left(\frac{3}{2},\frac{-3}{4},4\right)$
• B. $\left(1,2,\frac{-3}{5}\right)$
• C. $\left(\frac{3}{4},\frac{3}{2},\frac{-5}{3}\right)$
• D. $\left(\frac{-9}{2},\frac{1}{2},1\right)$

Answer 1

Answer: (C)

$\left(\frac{3}{4},\frac{3}{2},\frac{-5}{3}\right)$

$24x^2-14x^2-63x+45=0$

$\alpha ,2\alpha ,\beta$ let the roots

$3\alpha +\beta =\frac{7}{12}:\left(1\right)2{\alpha }^2\beta =\frac{-15}{8}$

$2{\alpha }^2+3\alpha \beta =\frac{-21}{8}...\left(2\right){\alpha }^2\beta =\frac{-15}{16}...(30$

From $\left(1\right),\beta =\frac{7}{12}-2\alpha$

such $\beta$ in $\left(2\right)$m we get

$2{\alpha }^2+3(\frac{7}{12}-3\alpha )=\frac{-21}{8}$

$2{\alpha }^2-9{\alpha }^2+\frac{7\alpha }{4}=\frac{-21}{8}$

$-7{\alpha }^2+\frac{7\alpha }{4}=\frac{-21}{8}$

$-{\alpha }^2+\frac{\alpha }{4}=\frac{-3}{8}$

$4{\alpha }^2+\frac{\alpha }{4}=\frac{-3}{8}$

$4{\alpha }^2-\alpha =\frac{3}{2}$

$8{\alpha }^2-2\alpha -3=0$

$8{\alpha }^2-6\alpha +4\alpha -3=0$

$2\alpha (4\alpha -3)+1(4\alpha -3)=0$

$\alpha =\frac{3}{4}$ or $\alpha =\frac{-1}{2}\Rightarrow \alpha =\frac{3}{4}$ is satisfied by given equation

Here $\beta =\frac{7}{12}-3\left(\frac{3}{4}\right)=\frac{7-27}{12}=\frac{-20}{12}=\frac{-5}{3}$

Hence Roots are $\frac{3}{4},\frac{3}{2}$ nd $\frac{-5}{3}$