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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