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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
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B
13
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C
23
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D
13
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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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