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Question

If $$ \alpha $$ and $$ \beta $$ be the zeroes of the polynomial $$ x^{2}+10 x+30, $$ then find the quadratic polynomial whose zeroes are $$ \alpha+2 \beta $$ and $$ 2 \alpha+\beta $$


Solution

Given: $$ \mathrm{p}(\mathrm{x})=\mathrm{x}^{2}+10 \mathrm{x}+30 $$

So, Sum of zeroes $$ =\alpha+\beta=\dfrac{-\mathrm{b}}{\mathrm{a}}=\dfrac{-10}{1}=-10 \ldots(1)$$

$$ =\alpha \beta=\dfrac{c}{a}=\dfrac{30}{1}=30 \ldots(2)$$

Product of zeroes Now, Let the zeroes of the quadratic polynomial be $$ \alpha^{\wedge}=\alpha+2 \beta, \beta^{\prime}=2 \alpha+\beta $$

Then, $$ a^{\prime}+\beta^{\prime}=a+2 \beta+2 a+\beta=3 \alpha+3 \beta=3(\alpha+\beta) $$

$$ \alpha^{\prime} \beta^{\prime}=(\alpha+2 \beta) \times(2 \alpha+\beta)=2 \alpha^{2}+2 \beta^{2}+5 \alpha \beta $$

Sum of zeroes $$ =3(\alpha+\beta) $$

Product of zeroes $$ =2 a^{2}+2 \beta^{2}+5 \alpha \beta $$

Then, the quadratic polynomial $$ =\mathrm{x}^{2}-(\text { sum of zeroes }) \mathrm{x}+ $$ product of zeroes $$ =x^{2}-(3(a+\beta)) x+2 a^{2}+2 \beta^{2}+5 a \beta $$

$$ =x^{2}-3(-10) x+2\left(a^{2}+\beta^{2}\right)+5(30)\left\{\text { from } e q^{n}(1) \&(2)\right\} $$

$$ =x^{2}+30 x+2\left(a^{2}+\beta^{2}+2 a \beta-2 a \beta\right)+150 $$

$$ =x^{2}+30 x+2(a+\beta)^{2}-4 a \beta+150 $$

$$ =x^{2}+30 x+2(-10)^{2}-4(30)+150 $$

$$ =x^{2}+30 x+200-120+150 $$

$$ =x^{2}+30 x+230 $$
So, the required quadratic polynomial is $$ x^{2}+30 x+230 $$

Mathematics

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