Question

- vulve any two of the following.
1) Find c, if the quadratic equation x? - 2 (c + 1)* + c = 0 has real and
equal roots.​

Answer 1

Question:

Find the value of c , if the quadratic equation x^2 - 2(c + 1)x + c^2 = 0 has real and equal roots.

Answer:

The required value of the "c" is -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;

x^2 - 2(c + 1)x + c^2 = 0.

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

=> D = [-2(c + 1)]^2 - 4•1•c^2

=> D = 4(c^2 + 2c + 1) - 4c^2

=> D = 4(c^2 + 2c + 1 - c^2)

=> D = 4(2c + 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(2c + 1) = 0

=> 2c + 1 = 0

=> 2c = -1

=> c = -1/2

Hence,

The required value of "c" is -1/2 .

Also,

Putting c = -1/2 , in the given quadratic equation, it will reduce to;

=> x^2 - 2(c + 1)x + c^2 = 0

=> x^2 - 2(-1/2 + 1)x + (-1/2)^2 = 0

=> x^2 - 2•(1/2)x + 1/4 = 0

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

=> (4x^2 - 4x + 1)/4 = 0

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

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

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

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

=> x = 1/2 , 1/2

Hence,

When c = -1/2 , then the equation will reduce to 4x^2 - 4x + 1 = 0 and its roots will be 1/2 , 1/2 .

Answer 2

Correct Question :

Find the value of c if the quadratic equation ${x}^{2}-2(c+1)x+{c}^{2}=0$ has real and equal roots.

Answer:

$\large\bold\red{c=-\frac{1}{2}}$

Step-by-step explanation:

Given,

a quadratic equation,

${x}^{2}-2(c+1)x+{c}^{2}=0$

Now, we know that,

general form of a quadratic equation is :-

$a {x}^{2} + bx + m = 0$

Comparing the coefficients ,

we get,

$a = 1 \\ \\ b = - 2(c + 1) \\ \\ m = {c}^{2}$

Now,

it is being given that,

the roots of quadratic equation are real and equal.

Therefore,

Descriminant, D = 0

$= > \sqrt{ {b}^{2} - 4am} = 0 \\ \\ = > {b}^{2} - 4am = 0 \\ \\ = > {b}^{2} = 4am$

Putting the values of a, b and m,

we get,

$= > {( - 2(c + 1))}^{2} = 4 \times 1 \times {c}^{2} \\ \\ = > 4 {(c + 1)}^{2} = 4 {c}^{2} \\ \\ = > {(c + 1)}^{2} = {c}^{2} \\ \\ = > {c}^{2} + 2c + 1 = {c}^{2} \\ \\ = > 2c + 1 = 0 \\ \\ = > c = - \frac{1}{2}$

Hence,

$\bold{c=-\frac{1}{2}}$