Question

find the value lf K if roots of given quadratic equations are real and equal Kx² + 4x + 4​

Answer 1

Answer: k = 1

Solution:

Given a quadratic equation kx² + 4x + 4 = 0 , whose roots are real and equal. We know,

• A quadratic equation has equal and real roots when its discriminant is equal to 0.

Also,

• Discriminant = ( coefficient of x )² - 4( coefficient of x² )( constant term )

Which means the discriminant of this quadratic equation is 0, now value of can be found in the following way...

> (4)² - 4 (k)(4) = 0

> 16 - 16k = 0

> -16k = -16

> 16k = 16

> k = 1

Hence, the value of k is 1.

Important properties:

• A quadratic equation has real and distinct roots if the value of its discriminant is greater than zero.

• A quadratic equation has real and equal roots if the value of its discriminant is equal to zero.

• A quadratic equation has imaginary roots if the value of its discriminant is less than zero.

Answer 2

Question:-

Find the value lf K if roots of given quadratic equations are real and equal Kx² + 4x + 4.

Solution:-

• Roots are real and equal.

Kx² + 4x + 4

ax² + bx + c

Given:-

• a = k,
• b = 4, and
• c = 4.

We know for roots to be real D ≥ 0.

D = b² – 4ac

16 – 4(4) k ≥ 0

16 – 16k ≥ 0

16 (1 – k) ≥ 0

1 – k ≥ 0

Multiply both – 1 both sides.

k – 1 ≤ 0

Answer:- k ≤ 1.

More Information:-

• Real and unequal,if b² – 4ac > 0.

• Rational and different,if b² – 4ac is a perfect square.

• Real and equal,if b² – 4ac = 0.

• If D < 0 roots are imaginary and unequal or complex conjugates.