Question

find the value of k so that the quadratic equation has equal roots (k+3)x^3+2(k+3)x+4=0​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ (k + 3)x² + 2(k + 3)x + 4 = 0.

As we know that,

For real and equal roots : D = 0.

⇒ b² - 4ac = 0.

⇒ [2(k + 3)²] - 4(k + 3)(4) = 0.

⇒ [4(k² + 9 + 6k) - 16(k + 3)] = 0.

⇒ 4k² + 36 + 24k - 16k - 48 = 0.

⇒ 4k² + 8k - 12 = 0.

⇒ 4(k² + 2k - 3) = 0.

⇒ k² + 2k - 3 = 0.

Factorizes the equation into middle term splits, we get.

⇒ k² + 3k - k - 3 = 0.

⇒ k(k + 3) - 1(k + 3) = 0.

⇒ (k - 1)(k + 3) = 0.

⇒ k = 1 and k = - 3.

                                                                                                                       

MORE INFORMATION.

Nature of the roots of the 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

Question:-

• Find the value of k so that the quadratic equation has equal roots ( k + 3 )x² + 2( k + 3)x + 4 = 0.

To Find:-

• Find the value of k.

Solution:-

Given ,

• x² coefficient = ( k + 3 )

• x coefficient = 2( k + 3 )

• constant term = 4

We have to find the value of k:-

$\dashrightarrow\sf \: { b }^{ 2 } - 4ac = 0$

$\dashrightarrow\sf \: { [ 2( k + 3 ) ] }^{ 2 } - 4( k + 3 )( 4 ) = 0$

$\dashrightarrow\sf \: 4( { k }^{ 2 } + 9 + 6k ) - 16k - 48 = 0$

$\dashrightarrow\sf \: { 4k }^{ 2 } + 24k + 36 - 16k - 48 = 0$

$\dashrightarrow\sf \: { 4k }^{ 2 } + 8k - 12 = 0$

$\dashrightarrow\sf \: 4( { k }^{ 2 } + 2k - 3 = 0$

$\dashrightarrow\sf \: { k }^{ 2 }+ 2k - 3 = 0$

$\dashrightarrow\sf \: { k }^{ 2 } + 3k - k - 3 = 0$

$\dashrightarrow\sf \: k( k + 3 ) - ( k + 3 ) = 0$

$\dashrightarrow\sf \: ( k + 3 ) ( k - 1 ) = 0$

$\dashrightarrow\sf \: k = 1 , -3$

Hence ,

• The value of k is 1 , -3