Question

if the difference of the zeries of the polynomial p(x)=4x^2-4kx+9 is 4,find the value of k
plzzzzz answer this fast​

Answer 1

GIVEN :–

• Difference of zeros of polynomial p(x) = 4x² - 4kx + 9 is 4.

TO FIND :–

• Value of 'k' = ?

SOLUTION :–

• We know that if a quadratic equation ax² + bx + c = 0 have two roots α & β , then Difference of roots –

$\\ \implies { \boxed{ \sf \:Difference \: \: of \: \: roots = \pm \: \dfrac{ \sqrt{D} }{a}}}$

$\\ \implies { \boxed{ \sf \:Difference \: \: of \: \: roots = \pm \: \dfrac{ \sqrt{ {b}^{2} - 4ac} }{a}}}$

• Here –

$\\ \: \: \: \: { \huge{.}} \: \: \sf \: a = 4$

$\\ \: \: \: \: { \huge{.}} \: \: \sf \: b= - 4k$

$\\ \: \: \: \: { \huge{.}} \: \: \sf \: c= 9$

• So that –

$\\ \implies \sf \:Difference \: \: of \: \: roots = \pm \: \dfrac{ \sqrt{ {b}^{2} - 4ac} }{a}$

$\\ \implies \sf \:4 = \pm \: \dfrac{ \sqrt{ {( - 4k)}^{2} - 4(4)(9)} }{4}$

$\\ \implies \sf \:4 = \pm \: \dfrac{ \sqrt{ 16 {k}^{2} - 4(36)} }{4}$

$\\ \implies \sf \:4 = \pm \: \dfrac{ \cancel4 \sqrt{ {k}^{2} - 9} } { \cancel4}$

$\\ \implies \sf \:4 = \pm \: \sqrt{ {k}^{2} - 9}$

• Square on both sides –

$\\ \implies \sf \:(4) ^{2} = \: { {k}^{2} - 9}$

$\\ \implies \sf \:16 = \: { {k}^{2} - 9}$

$\\ \implies \sf \: {k}^{2} = 25$

$\\ \implies \large { \boxed{ \sf \: k = \pm \sqrt5 }}$