Question

2] It is given that the difference between the zeros of 4x^2 -8kx+9 is 4 and k>0. Then k=?

Answer 1

GIVEN :–

• Difference between the zero's of 4x² - 8kx + 9 = 0 is 4.

• k > 0

TO FIND :–

• Value of k = ?

SOLUTION :–

• If a quadratic equation ax² + bx + c = 0 have two roots , then Difference of roots –

$\\ \bigstar \: \: { \red{ \boxed{ \bold{ Difference \: \: of \: \: \: roots \: = \dfrac{ \pm\sqrt{D} }{a} = \dfrac{ \pm \sqrt{ {b}^{2} - 4ac} }{a} \: \: }}}}$

• Here –

$\\ \: \: \: \: { \huge{.}}{ \blue{ \bold{ \: \: \: a = 4}}}$

$\\ \: \: \: \: { \huge{.}}{ \blue{ \bold{ \: \: \:b = - 8k}}}$

$\\ \: \: \: \: { \huge{.}}{ \blue{ \bold{ \: \: \:c = 9}}}$

• According to the question –

$\\ \implies { \bold{ Difference \: \: of \: \: \: roots \: = 4 }}$

$\\ \implies { \bold{ \dfrac{ \pm \sqrt{ {b}^{2} - 4ac} }{a} \: \: = 4 }}$

$\\ \implies { \bold{ \dfrac{ \pm \sqrt{ {( - 8k)}^{2} - 4(4)(9)} }{4} \: \: = 4 }}$

$\\ \implies { \bold{ \pm \sqrt{ {( - 8k)}^{2} - 4(4)(9) } \: \: = 16 }}$

• Square on both side –

$\\ \implies { \bold{ {( - 8k)}^{2} - 4(4)(9) \: \: = {(16)}^{2} }}$

$\\ \implies { \bold{ {( - 8k)}^{2} -144 \: \: = 256 }}$

$\\ \implies { \bold{ 64 {k}^{2} \: \: = 256 + 144 }}$

$\\ \implies { \bold{ 64 {k}^{2} \: \: = 400 }}$

$\\ \implies { \bold{ {k}^{2} \: \: = \dfrac{400}{64} }}$

$\\ \implies { \bold{ {k} \: \: = \pm\sqrt{\dfrac{400}{64} }} }$

$\\ \implies { \bold{ {k} \: \: = \pm{\dfrac{20}{8} }} }$

$\\ \implies { \bold{ {k} \: \: = \pm{\dfrac{5}{2} }} }$

• But k > 0 , So that –

$\\ \implies \large { \green{ \boxed{ \bold{ {k} \: \: = {\dfrac{5}{2} }} }}}$