Question

if zeros of a quadratic polynomial x²+(a+1)x+b are 2 and -3 then a=? b=? (ans using alfa beta formulas)​

Answer 1

EXPLANATION.

→ Zeroes of a quadratic polynomial

x² + ( a + 1 )x + b.

→ Zeroes are = 2 and -3.

→ Case = 1.

→ put the value of x = 2 in equation.

→ (2)² + ( a + 1)2 + b = 0

→ 4 + 2a + 2 + b = 0

→ 2a + b + 6 = 0

→ b = -2a - 6 ...........(1)

→ Case = 2.

→ Put the value of x = -3 in equation.

→ (-3)² + ( a + 1 )-3 + b = 0

→ 9 + [ -3a - 3 ] + b = 0

→ 9 - 3a - 3 + b = 0

→ 6 - 3a + b = 0 .........(2)

→ From equation (1) and (2) we get,

→ 6 - 3a + [ -2a - 6 ] = 0

→ 6 - 3a - 2a - 6 = 0

→ -5a = 0

→ a = 0

→ Put the value of a = 0 in equation (1)

we get,

→ b = -2(0) - 6

→ b = -6

→ Value of A = 0 and B = -6.

Answer 2

$\sf f \bigg( \: x \: \bigg) = {x}^{2} + \bigg(a + b \bigg)x + b \\ \\ \\ \sf f \bigg( \: 2 \: \bigg) = 0,f \bigg( - 3 \bigg) = 0 \\ \\ \\ \sf : \implies {2}^{2} + \bigg(a + 1 \bigg)2 + b = 0 \\ \\ \\ \sf : \implies \: 4 + 2a + 2 + b = 0 \\ \\ \\ \sf :\implies \: 2a + b = - 6 \\ \\ \\ \sf \bigg( - 3 { \bigg)}^{2} + \bigg(a + 1 \bigg) \bigg( - 3 \bigg) + b = 0 \\ \\ \\ \sf : \implies a - 3a - 3 + b = 0 \\ \\ \\ \sf : \implies - 3a + b = - 6 \\ \\ \\ \sf - 3ab + \cancel{b} = - 6 \\ \: \underline{ \sf \: \: 2a \: + \: \cancel{b} = - 6} \\ \sf \: \: \: \: \: - 5a = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: b = - 6 \\ \: \: \: \: \: \: \sf a = 0$