# Ex.4.4 Q2 Quadratic Equations Solutions - NCERT Maths Class 10

## Question

Find the value of \(k\) for each of the following quadratic equation , so that they have two equal roots.

(i) \(2x^\text{2}+kx+3=0\)

(ii) \(kx \left(x-2\right)+6=0\)

## Text Solution

**What is known?**

value of \(k.\)

**What is Known?**

Quadratic equation has equal real roots.

**Reasoning:**

Since the quadratic equation has equal real roots:

Discriminant \(b^\text{2}-4ac=0\)

**Steps:**

(i) \(2x^\text{2}+kx+3=0\)

\[a= 2,\;b = k,\;c = 3\]

\[\begin{align}{b^2} - 4ac &= 0\\{{(k)}^2} - 4(2)(3) &= 0\\{k^2} - 24 &= 0\\{k^2} &= 24\\k &= \sqrt {24} \\k &= \pm \sqrt {2 \times 2 \times 2 \times 3} \\k& = \pm 2\sqrt 6 \end{align}\]

(ii) \(kx \left(x-2\right)+6=0\)

\[a = k,\;b = - 2k,\;c = 6\]

\[\begin{align}{b^2} - 4ac &= 0\\{{( - 2k)}^2} - 4(k)(6) &= 0\\4{k^2} - 24k &= 0\\4k(k - 6) &= 0\\k = 6 & \qquad k = 0\\\end{align}\]

If we consider the value of *\(k\) *as \(0,\) then the equation will not longer be quadratic.

Therefore, \(k = 6\)