Question

Find the value of k for which the quadratic equation
4x2 — 2 (k + 1) x + (k + 4) = 0 has equal roots.

Answer 1

Question:

Find the value of k for which the quadratic equation 4x² - 2(k+1)x + (k+4) = 0 has equal roots.

Answer:

k = -3 , 5

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

4x² - 2(k+1)x + (k+4) = 0

Clearly , we have ;

a = 4

b = -2( k+1 )

c = k+4

We know that ,

The quadratic equation will have equal roots if its discriminant is equal to zero .

=> D = 0

=> [-2(k+1)]² - 4•4•(k+4) = 0

=> 4(k² + 2k + 1) - 4(4k+16) = 0

=> 4(k² + 2k + 1 - 4k - 16) = 0

=> k² - 2k - 15 = 0

=> k² - 5k + 3k - 15 = 0

=> k(k-5) + 3(k-5) = 0

=> (k-5)(k+3) = 0

=> k = -3 , 5

Hence,

The required values of k are -3 and 5 .

Answer 2

$\huge{\boxed{\red{\star\;Answer}}}$

$\large{\underline{\pink{\star\;Discriminent}}}$

If $ax^{2}+bx+c=0$ is a quadratic equation then

Discriminent is defined as follows

$D=b^{2}-4ac$

• If D > 0 , roots exist and they are real and distinct
• If D = 0 , roots exist and they are equal
• If D < 0 , roots are imaginery

Given equation , $4x^{2}+(2k+2)x+(k+4)=0$

Since given roots are equal,

$D={b^{2}-4ac=0}$

$D={{2k+2}^{2}-4(k+4)(4)=0}$

$4k^2+4+6k-16k-64=0$

$k^2-2k-15=0$

k = -3 or k = 5

$\large{\boxed{\green{The\;values\;of\;k\;are\;-3\;and\;5}}}$