Question

The roots of quadratic equation
kxw2+4x +1 = 0
of k=--
a) 4
b) -3 c) -4 d) -5
are equal and real, then the value
​

Answer 1

Correct Question:-

The roots of quadratic equation kx² + 4x + 1 = 0 are equal and real, then the value of k is ?

a) 4

b) -3

c) -4

d) -5

Concept:-

If the discriminant of the quadratic equation ax² + bx + c = 0, is equal to zero then the roots of the quadratic equation is equal and real.

Solution:-

Here, Equation = $\sf{kx^{2} + 4x + 1 = 0}$

And, a = k,         b = 4       and     c = 1

Now, Discriminant (D) ⇒ $\sf{b^{2} - 4ac}$

Putting the all values we get,

⇒ $\sf{4^{2} - 4 \times k \times 1}$

⇒ $\sf{16 - 4k}$

Since, It is given that roots of the quadratic equation are equal and real.

∴   $\sf{16 - 4k = 0}$

⇒ $\sf{4k = 16}$

⇒ $\sf{k = \dfrac{16}{4}}$

⇒ $\sf{k = 4}$

$\sf{Hence,\ The\ value\ of\ k\ in\ the\ equation\ is\ 4 .}\\\\\underline{\boxed{\tt{Option\ (a)\ is\ correct.}}}$

Answer 2

Correct Question:-

The roots of quadratic equation kx² + 4x + 1 = 0 are equal and real, then the value of k is ?

a) 4

b) -3

c) -4

d) -5

Concept:-

If the discriminant of the quadratic equation ax² + bx + c = 0, is equal to zero then the roots of the quadratic equation is equal and real.

Solution:-

Here, Equation = $\sf{kx^{2} + 4x + 1 = 0}$kx

2

+4x+1=0

And, a = k, b = 4 and c = 1

Now, Discriminant (D) ⇒ $\sf{b^{2} - 4ac}$b

2

−4ac

Putting the all values we get,

⇒ $\sf{4^{2} - 4 \times k \times 1}$4

2

−4×k×1

⇒ $\sf{16 - 4k}$16−4k

Since, It is given that roots of the quadratic equation are equal and real.

∴ $\sf{16 - 4k = 0}$16−4k=0

⇒ $\sf{4k = 16}4k=16$

⇒ $\sf{k = \dfrac{16}{4}}$

k = 416

⇒ $\sf{k = 4}k=4$

$\begin{lgathered}\sf{Hence,\ The\ value\ of\ k\ in\ the\ equation\ is\ 4 .}\\\\\underline{\boxed{\tt{Option\ (a)\ is\ correct.}}}\end{lgathered}$

Hence, The value of k in the equation is 4.

Option (a) is correct.