Question

find the value of k for which the roots are real and equal in equation:(k+1)x²-2(3k+1)x+8k+1=0 hence find the roots of the quadratic equation​

Answer 1

Answer:

The roots of the given quadratic equation are

$\displaystyle{\boxed{\red{\sf\:x\:=\:1}}\:\sf\qquad\:OR\:\qquad\:\boxed{\red{\sf\:x\:=\:\dfrac{5}{2}}}}$

Step-by-step-explanation:

The given quadratic equation is ( k + 1 ) x² - 2 ( 3k + 1 ) x + 8k + 1 = 0.

We have given that, the roots of the quadratic equation are real and equal.

We have to find the value of k and the roots of the quadratic equation.

Now,

( k + 1 ) x² - 2 ( 3k + 1 ) x + 8k + 1 = 0

Comparing with ax² + bx + c = 0, we get,

• a = ( k + 1 )
• b = - 2 ( 3k + 1 )
• c = 8k + 1

We know that,

For real and equal roots, discriminant of the quadratic equation is zero.

∴ b² - 4ac = 0

⇒ [ - 2 ( 3k + 1 ) ]² - 4 * ( k + 1 ) * ( 8k + 1 ) = 0

⇒ ( - 6k - 2 )² - ( 4k + 4 ) * ( 8k + 1 ) = 0

⇒ ( - 6k )² - 2 * ( - 6k ) * 2 + ( - 2 )² - [ 4k ( 8k + 1 ) + 4 ( 8k + 1 ) ] = 0

⇒ 36k² + 24k + 4 - ( 32k² + 4k + 32k + 4 ) = 0

⇒ 36k² + 24k + 4 - 32k² - 4k - 32k - 4 = 0

⇒ 36k² - 32k² + 24k - 4k - 32k + 4 - 4 = 0

⇒ 4k² + 20k - 32k = 0

⇒ 4k² - 12k = 0

⇒ 4k ( k - 3 ) = 0

⇒ 4k = 0 $\qquad\sf\:OR\:\qquad$ ( k - 3 ) = 0

⇒ k = 0 $\qquad\sf\:OR\:\qquad$ k - 3 = 0

⇒ k = 0 $\qquad\sf\:OR\:\qquad$ k = 3

∴ The value of k is 0 or 3.

─────────────────────────

Now, by substituting k = 0 in the given equation, we get,

( k + 1 ) x² - 2 ( 3k + 1 ) x + 8k + 1 = 0

⇒ ( 0 + 1 ) x² - 2 ( 3 * 0 + 1 ) x + 8 * 0 + 1 = 0

⇒ x² - 2 ( 0 + 1 ) x + 1 = 0

⇒ x² - 2x + 1 = 0

⇒ x² - x - x + 1 = 0

⇒ x ( x - 1 ) - 1 ( x - 1 ) = 0

⇒ ( x - 1 ) ( x - 1 ) = 0

⇒ ( x - 1 ) = 0

⇒ x - 1 = 0

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:x\:=\:1}}}}$

─────────────────────────

Now, by substituting k = 3 in the given equation, we get,

( k + 1 ) x² - 2 ( 3k + 1 ) x + 8k + 1 = 0

⇒ ( 3 + 1 ) x² - 2 ( 3 * 3 + 1 ) x + 8 * 3 + 1 = 0

⇒ 4x² - 2 ( 9 + 1 ) x + 24 + 1 = 0

⇒ 4x² - ( 2 * 10 ) x + 25 = 0

⇒ 4x² - 20x + 25 = 0

⇒ 4x² - 10x - 10x + 25 = 0

⇒ 2x ( 2x - 5 ) - 5 ( 2x - 5 ) = 0

⇒ ( 2x - 5 ) ( 2x - 5 ) = 0

⇒ ( 2x - 5 ) = 0

⇒ 2x - 5 = 0

⇒ 2x = 5

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:x\:=\:\dfrac{5}{2}}}}}$

∴ The roots of the given quadratic equation are

$\displaystyle{\boxed{\red{\sf\:x\:=\:1}}\:\sf\qquad\:OR\:\qquad\:\boxed{\red{\sf\:x\:=\:\dfrac{5}{2}}}}$