Question

find the value of k if the roots of the quadratic equation is 2 x square - 6 x + k equal to zero are real and equal​

Answer 1

Answer

Given :-

Quadratic equation - $\sf 2x^2 - 6x + k = 0$

Roots of quadratic equation are real and equal.

To Find :-

Value of k

Solution :-

When the roots of the quadratic equation are real and equal, the value of discriminant (D) is equal to 0.

Quadratic equation - $\sf 2x^2 - 6x + k = 0$

• a = 2
• b = -6
• c = k

Substituting the values :-

⟹ $\sf D = b^2 - 4ac = 0$

⟹ $\sf (-6)^2 - 4 \times 2 \times k = 0$

⟹ $\sf 36 - 8k = 0$

⟹ $\sf 8k = 36$

⟹ $\sf k = \frac{\cancel 36}{\cancel 8}$

⟹ $\boxed{\sf k = \frac{9}{2}}$

Additional information :-

• When the roots are real and distinct :-

D > 0 i.e. b² - 4ac > 0

• When the roots are imaginary and distinct :-

D < 0 i.e. b² - 4ac < 0

Answer 2

Question:-

Find the value of k if the roots of the quadratic equation is 2x² - 6x + k = 0 are real and equal.

Solution:-

The roots of the quadratic equation are real and equal.

Now,

We have to find the value of " k "

=> a = 2 ; b = -6 ; c = k

=> b² - 4ac = 0

=> (-6)² - 4(2)(k) = 0

=> 36 - 8k = 0

=> 36 = 8k

=> k = 36/8

=> k = 9/2

Therefore:-

$\red{\boxed{The \: value \: of \: k \: = \frac{9}{2} }}$

Verification:-

=> b² - 4ac = 0

=> (-6)² - 4(2)(9/2) = 0

=> 36 - 36 = 0

=> 0 = 0