Question

Find the values of k for which the quadratic equation (k + 4)x2 + (k + 1)x + 1 = 0 has equal roots. Also find these roots.

Answer 1

EXPLANATION.

Quadratic equation.

⇒ (k + 4)x² + (k + 1)x + 1 = 0.

As we know that,

D = Discriminant.

Roots are real and equal.

⇒ D = 0 Or b² - 4ac = 0.

⇒ (k + 1)² - 4(k + 4)(1) = 0.

⇒ (k² + 1 + 2k) - (4k + 16) = 0.

⇒ k² + 1 + 2k - 4k - 16 = 0.

⇒ k² - 2k - 15 = 0.

As we know that,

Factorizes into middle term split, we get.

⇒ k² - 5k + 3k - 15 = 0.

⇒ k(k - 5) + 3(k - 5) = 0.

⇒ (k + 3)(k - 5) = 0.

⇒ k = -3  and  k = 5.

                                                                                                                         

MORE INFORMATION.

Conditions for common roots.

Let quadratic equations are a₁x² + b₁x + c₁ = 0  and  a₂x² + b₂x + c₂ = 0.

(1) = If only one root is common.

x = b₁c₂ - b₂c₁/a₁b₂ - a₂b₁.

y = c₁a₂ - c₂a₁/a₁b₂ - a₂b₁.

(2) = If both roots are common.

a₁/a₂ = b₁/b₂ = c₁/c₂.

Answer 2

Question:-

• Find the values of k for which the quadratic equation (k + 4)x² + (k + 1)x + 1 = 0 has equal roots. Also find these roots.

To Find:-

• Find these roots.

Given:-

• (k + 4)x2 + (k + 1)x + 1 = 0 has equal roots.

Solution:-

Delta = b² - 4ac

$\tt\implies \: { ( k + 1 ) }^{ 2 } - 4( k + 4 )( 1 ) = 0$

$\tt\implies \: ( { k }^{ 2 } + 1 + 2k ) - ( 4k + 16 ) = 0$

$\tt\implies \: { k }^{ 2 } + 1 + 2k - 4k - 16 = 0$

$\tt\implies \: { k }^{ 2 } - 2k - 15 = 0$

Now ,

We have to find the roots by factorisation:

$\tt\implies \: { k }^{ 2 } - 5k + 3k - 15 = 0$

$\tt\implies \: k( k - 5 ) + 3( k - 5 ) = 0$

$\tt\implies \: ( k - 5 ) ( k + 3 ) = 0$

$\tt\implies \: k = 5 ; - 3$