Question

(iii) (k - 3)x2 +4(k - 3)x + 4 = 0
find the value of k for which the roots real and equation​

Answer 1

Given

$equation \: is \: (k - 3) {x}^{2} + 4(k - 3)x + 4 = 0$

solving the equation further

then it becomes

$(k - 3) {x}^{2} +( 4k - 12)x + 4 = 0$

To find

the real values of k

Method

Given the values of k are real

$\sf hence \:determinant\: should\: be \:greater\\ \sf than \:or \: equal \:to \:zero$

$\sf determinant\geqslant0$

$\sf determinant =b^2-4ac \geqslant0$

$\sf \bold{where \:b \:is\: coefficient\: of\: x}\\ \sf\bold{a\: is\: coefficient \:of\: x^2 }\\\sf\bold{c \:is\: the\: constant \:term}$

here in the equation

$b = (4k - 12) \\ a = (k - 3) \\ c = 4$

$= > {b}^{2} - 4ac \geqslant 0 \\ consider \: that \: it \: is \: equal \: to \: 0$

$= > {(4k - 12)}^{2} - 4(k - 3)(4) = 0$

expanding using

$\sf\underline{(a-b)^2=a^2+b^2-2ab}$

$= > 16 {k}^{2} + 144 - 96k - 16k - 48 = 0$

$= > 16 {k}^{2} - 112k + 96 = 0$

$k=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$k = \dfrac{112 + \sqrt{12544 - 6144} }{32} \\and \\ k = \dfrac{112 - \sqrt{12544 - 6144} }{32}$

$k= \dfrac{112\pm\sqrt{6400}}{32}$

$k=\dfrac{112\pm80}{32}$

$k=\dfrac{192}{32},k=\dfrac{32}{32}$

$\bf\boxed{k=6}$

and

$\bf\boxed{k=1}$

are the real roots of the given equation