Question

Find the value of K, if the equation 2^2 + k + 2 = 0 has equal roots.

Answer 1

d=b^2-4ac solve from this formula

Answer 2

Answer:-

The Required Value of k is 4.

Explanation:-

Given:-

• A quadratic equation $\bf 2x^2+k+2=0.$

• The roots of the given equation is equal.

To Find:-

• The value of k.

Concepts Used:-

In a quadratic equation or polynomial if,

$\\ \bf\bullet {b}^{2} - 4ac = 0. \\ \\ \rm \: \: the \: \: roots\: would \: \: be \: \: real \: \: and \: \: equal.$

$\\ \bf\bullet {b}^{2} - 4ac > 0. \\ \\ \rm \: \: the \: \: roots \: would \: \: be \: \: real \: \: and \: \: distict.$

$\\ \bf\bullet {b}^{2} - 4ac < 0. \\ \\ \rm \: \: no \: \: real \: \: roots \: \: would \: \: be \: \: possible.$

Where,

• a = Coefficient of x².

• b = Coefficient of x.

• c = Constant Term.

So Here,

• a = 2.

• b = k.

• c = 2

And the zeroes are real and equal,

Therefore,

$\\ \tt\mapsto {b}^{2} - 4ac = 0.$

$\\ \tt\mapsto(k) {}^{2} - 4(2)(2) = 0. \\ \\ \rm by \: \: putti ng \: \: values.$

$\\ \tt\mapsto {k}^{2} - 4(4) = 0.$

$\\ \tt\mapsto {k}^{2} - 16 = 0.$

$\\ \tt\mapsto {k}^{2} = 0 + 16.$

$\\ \tt\mapsto {k}^{2} = 16.$

$\\ \tt\mapsto k = \sqrt{16} .$

$\\ \tt\mapsto k = \sqrt{4 \times 4} .$

$\\ \large\red{\mapsto\boxed{\blue{\bf k = 4}}}.$

Therefore The Required Value of k is 4.