Question

find the value of k for which of the which of the following quadratic equation have equal root (3k+1) x^2+2(k+1)x+1=0​

Answer 1

Question:

Find the value of k for which the quadratic equation (3k+1)x² + 2(k+1)x + 1 = 0 has equal roots.

Answer:

k = 0 , 1

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 ;

(3k+1)x² + 2(k+1)x + 1 = 0

Clearly , we have ;

a = 3k+1

b = 2(k+1)

c = 1

We know that ,

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

=> D = 0

=> [2(k+1)]² - 4•(3k+1)•1 = 0

=> 4(k²+2k+1) - 4(3k+1) = 0

=> 4[k² + 2k +1 - 3k - 1] = 0

=> k² - k = 0

=> k(k-1) = 0

=> k = 0 , 1

Hence,

Hence,The required values of k are 0 and 1 .

Answer 2

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

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

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

Discriminent is defined as follows

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

• 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 , $(3k+1)x^{2}+(2k+2)x+1=0$

Since given roots are equal,

• $D={b^{2}-4ac=0}$
• $D={{2k+2}^{2}-4(3k+1)=0}$
• ${(2k+2)}^{2}-4(3k+1)=0$
• $4k^2+4+6k-12k-4=0$
• $4(k^2-k)=0$
• k = 0 or k = 1

$\huge{\boxed{\green{The\;values\;of\;k\;are\;0\;and\;1}}}$