Question

Find the value of k for which the quadratic equation
(k + 1)x2 -6(k + 1)x + 3(k + 9) = 0, k# - 1 has equal roots.​

Answer 1

Given,

A quadratic equation (k + 1)x² -6(k + 1)x + 3(k + 9) = 0

To Find,

The value of k for which this equation has equal roots.

Solution,

The condition for a quadratic equation to have equal roots is

D = b²-4ac = 0

where,

b is the coefficient of x

a is the coefficient of x²

c is the constant

(-6(k+1))² = 4(k+1)(3(k+9)

36(k+1) = 12(k+9)

3k+3 = k+9

2k = 6

k = 3

Hence, for k = 3 this equation has equal roots.

Answer 2

Value of k=3, for quadratic equation to be equal roots.

Given:

• A quadratic equation.
• $(k + 1) {x}^{2} - 6(k + 1)x + 3(k + 9) = 0, \: k \neq - 1$

To find:

• Find the value of k, if given equation has equal roots.

Solution:

Concept to be used:

• If a quadratic equation $\bf a {x}^{2} + bx + c = 0, \: a \neq0$ have equal roots then it's discriminate (D) = 0.
• $\bf D = {b}^{2} - 4ac$

Step 1:

Compare the given equation with standard equation and write the values of a,b and c.

It is clear that

$a = k + 1$

$b = - 6(k + 1)$

and

$c = 3(k + 9)$

Step 2:

Put the values in D and solve for k.

As equation have equal roots if

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

So,

${( - 6(k + 1)})^{2} - 4(k + 1)(3(k + 9)) = 0$

or

$36(k + 1 )^{2} - 12(k + 1)(k + 9) = 0$

or

take 12(k+1) common from both terms.

$12(k + 1)(3(k + 1) - (k + 9) )= 0$

or

$12(k + 1) = 0$

or

$k + 1 = 0$

or

$\bf \red{k = - 1}$

but it is given that k≠-1.

So, we can't take this value.

Or

$3k + 3 - k - 9 = 0$

or

$2k - 6 = 0$

or

$2k = 6$

or

$\bf \green{k = 3}$

Thus,

Value of k=3, for quadratic equation to be equal roots.

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