Question

The value of k for which the equation has real roots is
$kx {}^{2} + 4x + 1 = 0$
A)
$k \geqslant 4$
B)
$k \leqslant 4$
C)
$k = 4$
D)
$k \leqslant - 4$
​

Answer 1

EXPLANATION.

Value of k for which equation has real roots.

Equation : kx² + 4x + 1 = 0.

As we know that,

For real roots : D ≥ 0.

⇒ b² - 4ac ≥ 0.

⇒ (4)² - 4(k)(1) ≥ 0.

⇒ 16 - 4k ≥ 0.

⇒ 4k - 16 ≤ 0.

⇒ 4k ≤ 16.

⇒ k ≤ 4.

Option [B] is correct answer.

                                                                                                                         

MORE INFORMATION.

Nature of the roots of quadratic expression.

(1) Real and unequal, 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

Step-by-step explanation:

given :

• The value of k for which the equation has real roots is kx2 + 4x +1 = 0

to find :

• kx2 + 4x +1 = 0

solution :

first we have :

• a = k , b = 4 or C = 1

• discriminant = b² - 4ac

we have value :

• a = k b= 4 and c = 1

• discriminant = (4)²-4(k)(1)

• discriminant = 16 - 4k

then ,we can solving equation :

• 16-4k = 0

• 4k = 16

• k = 16/4

• k = 4

option B is correct

learn more :

D = b² - 4ac is also known as discriminant

in any quardtic equation we use one formula because the solution comes :

x = - b ± √b² - 4ac / 2a