Question

# The value of $$a$$ for which one root of the quadratic equation $$\displaystyle (a^{2}-5a+3)x^{2}+(3a-1)x+2= 0$$ is twice as large as the other is

A
23
B
13
C
23
D
13

Solution

## The correct option is A $$\displaystyle \frac{2}{3}$$Given equation $$\displaystyle (a^{2}-5a+3)x^{2}+(3a-1)x+2= 0$$ Let $$\alpha$$ be the root of the equation.Hence, other root will be $$2\alpha$$.Now, $$3\alpha=\dfrac{1-3a}{a^{2}-5a+3}$$and $$2\alpha^{2}=\dfrac{2}{a^{2}-5a+3}$$$$\alpha^{2}=\dfrac{1}{a^{2}-5a+3}$$Hence,$$\dfrac{(1-3a)^{2}}{9(a^{2}-5a+3)^{2}}=\dfrac{1}{a^{2}-5a+3}$$$$9(a^{2}-5a+3)=9a^{2}-6a+1$$$$-45a+27=-6a+1$$$$26=39a$$$$a=\dfrac{2}{3}$$.Mathematics

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