1.

If the zeroes of the quadratic polynomial x²+(a+1)x+ bare 2 and -3, then find the value of a and b.​

Answer»

\huge\sf \red{Solution:-}

Let 2 zeroes be a and b of POLYNOMIAL,

x² + px + q = 0

sum of roots = a + b = -p/1 = -p products of roots = ab = q/1 = q

(a + b)² = a² + b² + 2ab = a² + b² = p² - 2Q

(a-b)² = a² + b² -2ab = p² - 2q -2q = p² -4q

now the question asks for new QUADRATIC EQ whose roots are (a+b) ² and (a-b)² 2

so sum of new roots are = (a+b)² + (a-b)²

so sum of new roots are = (a+b)² + (a-b)² =p² + p² -4q = 2p² - 4q

and product of roots = (a+b)²(a-b)² = 4 (p²)² (p²-4q)² = pª (p² +16q² + 8p²q)

hence new quadratic eq gonna be = x²-x(sum of roots) + (products of roots)

x²-x(2p² - 4q) + pª(p4 + 16q² +8p²q) = 0


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