Question

Obtain a quadratic polynomial whose zeroes are 2α + β and α + 2β where α and β are the zeroes of x2 + 3x – 10.​

Answer 1

Answer:

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Step by step solution:

$Given \ p(x)=x^{2}+3x-10\\\\First \ let \ us \ solve\\\\p(x)=x^{2}+3x-10\\\\p(x)=x^{2}+5x-2x-10\\\\p(x)=x(x+5)-2(x+5)\\\\p(x)=(x+5)(x-2)\\\\\alpha=-5 \ and \ \beta=2\\ \\ putting \ in \ given \ zeroes\\\\2\alpha +\beta =-10+2=-8\\\\\alpha+2\beta=-5+4=-1\\ \\their \ sum=-8-1=-9 \ and \ their \ product=-8\times-1=8\\\\Required \ polynomial=x^{2}-(\alpha+\beta)x+\alpha \beta\\\\Required \ polynomial=x^{2}+9x+8$

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