Question

Find the value of k where the equation has equal roots: (k-4)x^2 +2 (k-4)x+4

Answer 1

Answer:

$\mathfrak{\large{\underline{\underline{Answer}}}} = \bold{k = 8} \\ \bold{k = 4}$

Step-by-step explanation:

$(k - 4) {x}^{2} + 2(k - 4)x + 4 = 0 \\\bold{d = {b}^{2} - 4ac = 0 } \\ 2(k - 4) {}^{2} - 4(k - 4)(4) = 0 \\ 4( {k}^{2} + 16 - 8k) - 4(4k -1 6) = 0 \\ 4 {k}^{2} + 64 - 32k - 16k + 64 = 0 \\ 4 {k}^{2} - 48k + 128 = 0 \\ {k}^{2} - 12k + 32 = 0 \\ {k}^{2} - 8k - 4k + 32 = 0 \\ k(k - 8) - 4(k - 8) = 0 \\ (k - 4)(k - 8) = 0 \\\bold{k = 4} \\ \bold{k = 8}$

Answer 2

QUESTION:

Find the value of k where the equation has equal roots: (k-4)x^2 +2 (k-4)x+4

ANSWER:

Given equation has equal roots it means that discriminant is 0.

We use the discriminant formula to find the value of k.

$\huge\orange {d = {b}^{2} - 4ac}$

where;

b = coefficient of x

a = coefficient of x square

c = constant term

now come to main question;

Here;

$\red {(k - 4) {x}^{2} + 2(k - 4)x + 4 = 0 \\ (k - 4) {x}^{2} + (2k - 8)x + 4 = 0}$

b = 2k-8

c = 4

a = k-4

using the formula;

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

$0 = ( {2k - 8})^{2} - 4 \times (k - 4) \times 4 \\ 0 = {(2k)}^{2} - 2 \times 2k \times 8 + {8}^{2} - 4(4k - 16) \\ 4 {k}^{2} - 32k + 64 - 16k +64 = 0 \\ 4 {k}^{2} -48k+ 128 = 0$

$\red {(taking \: 4 \: common)}$

$4( {k}^{2} - 12k + 32) = 0$

$\huge\orange { {k}^{2} - 12k + 32 = 0}$

as it is a quadratic equation we have to factorised it,

Splitting the middle term.

In such a way that it's sum equal to -12 and product equal to 32.

${k}^{2} - 8k - 4k + 32 = 0 \\ k(k -8 ) - 4(k - 8) = 0$

$\huge\blue {(k - 4)(k - 8) = 0}$

so,

$k - 4 = 0 \\ k = 4 \\ \\ or \\ \\ k - 8 \: = 0 \\ k = 8$

FINAL ANSWER :

$\huge\pink {k = \: 4 \: or \: 8}$