Question

Find the zeroes of the quardratic equation 4 x^2-4x1 and
verify the relationship between the zeroes
and the
Cofficients.​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ 4x² - 4x + 1 = 0.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-4)/4 = 1.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 1/4.

As we know that,

Factorizes the equation into middle term splits, we get.

⇒ 4x² - 4x + 1 = 0.

⇒ 4x² - 2x - 2x + 1 = 0.

⇒ 2x(2x - 1) - 1(2x - 1) = 0.

⇒ (2x - 1)(2x - 1) = 0.

⇒ (2x - 1)² = 0.

⇒ 2x - 1 = 0.

⇒ x = 1/2.

Sum of the value of x :

⇒ x = 1/2 + 1/2.

⇒ x = 2/2 = 1.

Products of the value of x :

⇒ x = 1/2 x 1/2.

⇒ x = 1/4.

Hence verified.

                                                                                                                                         

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex conjugate.

Answer 2

Given :-

$\sf 4x^{2} -4x + 1 = 0$

To find :-

Relationship between the zeroes  and the  Coefficients.​

Solution :-

We are knowing that

α + β = -b/a

$\sf \alpha +\beta = \dfrac{-(-4)}{1}$

$\sf \alpha +\beta = \dfrac{4}{1}$

$\alpha +\beta =4$

Product of zeroes

$\sf\alpha \beta =\dfrac{c}{a}$

$\sf \alpha \beta =\dfrac{1}{4}$

On factorizing

4x² - 4x + 1 = 0

$\sf 4x^{2} -(2x + 2x) - 1 = 0$

$\sf 4x^{2} - 2x -2x +1 = 0$

$\sf 2x(2x - 1) - 1(2x - 1) = 0.$

Taking 2x - 1 as common

$\sf (2x - 1)^2 = 0.$

$\sf x = 1/2$

Sum

x = 1/2 + 1/2

x = 1 + 1/2

x = 2/2

x = 1/1

x = 1

Product of zeroes

x = 1/2 x 1/2.

x = 1 x 1/2 x 2

x = 1/4