Question

Q.27Find the zeros of the quadratic polynomial x2 +7x+12 and verify relation between the
zeros and its coefficients.​

Answer 1

★ AnswEr :-

$\implies \boxed{ \red{ \sf{x = (- 4) \: and \: (- 3)}}}$

★ GivEn :-

• The quadratic polynomial x^2 + 7x + 12 = 0

★ To Find :-

• The zeroes of the quadratic polynomial
• The relation between the zeroes and co efficients

★ Solution :-

Normal Equation of Quadratic Polynomial is,

$\boxed{a {x}^{2} + bx + c = 0 \: (a \neq \: 0)}$

Factorising the polynomial we get,

$\sf \: {x}^{2} + 7x + 12 = 0$

$\implies \: \sf \: {x}^{2} + (4 + 3)x + 12 = 0$

$\implies \: \sf \: {x}^{2} + 4x + 3x + 12 = 0$

$\implies \: \sf \: x(x + 4) + 3(x + 4)= 0$

$\implies \: \sf \: (x + 4) (x + 3)= 0$

$\implies \sf \: (x + 4) = 0$

$\implies \sf \: (x + 3) = 0$

$\implies \sf \: x = ( - 4)$

$\implies \sf \: x = ( - 3)$

$\implies \boxed{ \red{ \sf{x = (- 4) \: and \: (- 3)}}}$

The zeroes are -4 and -3.

And the co efficients of x^2, x^1 and x^0 are 1, 7, 12 respectively.

The sum of the zeroes (-4) + (-3) = (-7) = (-7/1)

$\sf \: Sum \: of \: zeroes = - \dfrac{b}{a}$

The product of the zeroes -4 × -3 = 12

$\sf \: Product \: of \: zeroes = \dfrac{c}{a}$

Hence, verified