Question

find the zeroes of the polynomial q(x)=4x²-4x-3 and verify the relationship between its zeroes and coefficient. ​

Answer 1

S O L U T I O N :

We have quadratic polynomial q(x) = 4x² - 4x - 3 & zero of the polynomial q(x) = 0.

By using factorization method :

⇒ 4x² - 4x - 3 = 0

⇒ 4x² - 6x + 2x - 3 = 0

⇒ 2x(2x - 3) + 1(3x - 3) = 0

⇒ (2x - 3) (2x + 1) = 0

⇒ 2x - 3 = 0  Or  2x + 1 = 0

⇒ 2x = 3  Or  2x = -1

⇒ x = 3/2  Or  x = -1/2

∴ α = 3/2 & β = -1/2 are two zeroes of the given polynomials .

Now,

As we know that quadratic polynomial compared with ax² + bx + c;

• a = 4
• b = -4
• c = -3

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha +\beta = \dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\bigg\rgroup}$

$\mapsto\tt{\dfrac{3}{2} +\bigg(-\dfrac{1}{2} \bigg) = \dfrac{-(-4)}{4}}$

$\mapsto\tt{\dfrac{3}{2} -\dfrac{1}{2} = \dfrac{4}{4}}$

$\mapsto\tt{\dfrac{3-1}{2} = \dfrac{4}{4}}$

$\mapsto\tt{\cancel{\dfrac{2}{2} } = \cancel{\dfrac{4}{4}}}$

$\mapsto\bf{1=1}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}\bigg\rgroup}$

$\mapsto\tt{\dfrac{3}{2} \times \bigg(-\dfrac{1}{2} \bigg) = \dfrac{-3}{4}}$

$\mapsto\bf{\dfrac{-3}{4} = \dfrac{-3}{4}}$

Thus,

The relation between zeroes & coefficient are verified .