Question

find the values of k for which quadratic equation (3k+1)x2+2( k+1)x+1=0 has equal roots also find the roots​

Answer 1

Question:

Find the values of k for which quadratic quadratic (3k+1)x^2 + 2(k+1)x + 1 = 0 has equal roots , also find the roots.

Answer:

• k = 0 or 1.

• If k = 0, then the roots of the equation will be -1, -1.

• If k = 1, then the roots of the equation will be -1/2 , -1/2 .

Note:

• The degree of an equation decides the number of its roots.

• The maximum number of roots of an equation ie equal to the degree of that equation.

• The degree of a quadratic equation is two , thus the maximum number of roots of s quadratic equation is two.

• If we have a quadratic equation say; ax^2 + bx + c = 0 , then the determinant of the equation is given by ;

D = b^2 - 4•a•c .

• If the determinant of the quadratic equation is greater than zero (ie; D > 0), then its roots are real and distinct.

• If the determinant of the quadratic equation is equal to zero (ie; D = 0) , then its roots are real and equal.

• If the determinant of the quadratic equation is less than zero (ie; D < 0) , then its roots are imaginary (not real).

Solution:

Here ,

The given quadratic equation is;

(3k+1)x^2 + 2(k+1)x + 1 = 0.

The determinant for the given quadratic equation will be given as;

=> D = [2(k+1)]^2 - 4•(3k+1)•1

=> D = 4(k^2 + 2k + 1) - 4(3k + 1)

=> D = 4(k^2 + 2k + 1 - 3k - 1)

=> D = 4(k^2 - k)

=> D = 4•k•(k - 1)

Also,

We know that,

For real and equal roots of a quadratic equation, its determinant must be equal to zero.

ie;

=> D = 0

=> 4•k•(k - 1) = 0

=> k•(k - 1) = 0

=> k = 0 , 1

Hence,

The required values of "k" are ;

0 , 1 .

Case(1)

When k = 0 , then the equation will be;

=> (3k+1)x^2 + 2(k+1)x + 1 = 0

=> (3•0+1)x^2 + 2(0+1)x + 1 = 0

=> x^2 + 2x + 1 = 0

=> x^2 + x + x + 1 = 0

=> x(x + 1) + 1(x + 1) = 0

=> (x + 1)(x + 1) = 0

=> x = -1 , -1

Case(2)

When k = 1, then the equation will be;

=> (3k+1)x^2 + 2(k+1)x + 1 = 0

=> (3•1+1)x^2 + 2(1+1)x + 1 = 0

=> 4x^2 + 4x + 1 = 0

=> 4x^2 + 2x + 2x + 1 = 0

=> 2x(2x + 1) + 1(2x + 1) = 0

=> (2x + 1)(2x + 1) = 0

=> x = -1/2 , -1/2

Hence,

When k = 0 , then the roots of the equation will be -1 , -1 .

When k = 1, then the roots of the equation will be -1/2 , -1/2 .

Answer 2

Answer:

0 and 1 are the required values of k.

Step-by-step explanation:

Given :

Quadratic equation = $\sf (3k+1)x2+2( k+1)x+1=0$

To Find :

The values of k.

Solution :

As we know,

For equal roots, D = 0.

So,

$\sf\\ \\\implies b^2-4ac=0 [2(k+1)]^2-4(3k+1)\times1=0\\ \\ \\ \implies 4(k^2+1+2k)-12k-4=0\\ \\ \\ \implies 4(k^2+1+2k-3k-1)=0\\ \\ \\\implies k^2+1+2k-3k-1=0\\ \\ \\ \implies k^2-k=0\\ \\ \\ \implies k(k-1)=0\\ \\ \\ \implies k=0\quad or\quad k-1=0$

Since, k can't be 0 again.

So,

=> k - 1 = 0

=> k = 1

∴ 0 and 1 are the required values of k.

$\rule{300}{1.5}$

Finding roots.

• When k = 0.

$\sf\\ \\\implies (3k+1)x^2 + 2(k+1)x + 1 = 0\\ \\ \\\implies (3\times0+1)x^2 + 2(0+1)x + 1 = 0\\ \\ \\\implies  x^2 + 2x + 1 = 0\\ \\ \\\implies x^2 + x + x + 1 = 0\\ \\ \\\implies x(x + 1) + 1(x + 1) = 0\\ \\ \\\implies  (x + 1)(x + 1) = 0\\ \\ \\\implies \bf  x = -1 , -1$

• When k = 1.

$\sf\\ \\\implies (3k+1)x^2 + 2(k+1)x + 1 = 0\\ \\ \\\implies (3\times1+1)x^2 + 2(1+1)x + 1 = 0\\ \\ \\\implies 4x^2 + 4x + 1 = 0\\ \\ \\\implies 4x^2 + 2x + 2x + 1 = 0\\ \\ \\\implies 2x(2x + 1) + 1(2x + 1) = 0\\ \\ \\ \implies (2x + 1)(2x + 1) = 0\\ \\ \\\implies \bf x = \dfrac{-1}{2} , \dfrac{-1}{2}$