Question

For what value of k the equation 2kx2 - 2(1 + 2k)x + (3 + 2k) have real but distinct roots. When will the roots be equal?

Answer 1

GIVEN :–

• A quadratic equation 2kx² -2 (1 + 2k) x + (3 + 2k) = 0 have real but distinct roots.

TO FIND :–

• Value of 'k' = ?

• Value of 'k' when roots be equal .

SOLUTION :–

• If a quadratic equation have real and Distinct roots then discriminate will be –

$\\ \longrightarrow \: \large{ \boxed{ \sf D > 0 }}$

$\\ \longrightarrow \: \large{ \boxed{ \sf {b}^{2} - 4ac > 0 }}$

• Here –

$\\ \: \: \: \: \: \: { \huge{.}} \: \: \sf \: a = 2k$

$\\ \: \: \: \: \: \: { \huge{.}} \: \: \sf \: b = - 2(1 + 2k)$

$\\ \: \: \: \: \: \: { \huge{.}} \: \: \sf \: c= 3 + 2k$

• So that –

$\\ \implies \: \sf {b}^{2} - 4ac > 0$

$\\ \implies \: \sf { \{ - 2(1 + 2k)} \}^{2} - 4(2k)(3 + 2k) > 0$

$\\ \implies \: \sf { 4(1 + 2k)}^{2} - 4(6k + 4{k}^{2} ) > 0$

$\\ \implies \: \sf { 4(1 + 2k)}^{2} - 24k - 16 {k}^{2} > 0$

$\\ \implies \: \sf { 4(1 + 4 {k}^{2} + 4k)}- 24k - 16 {k}^{2} > 0$

$\\ \implies \: \sf 4 + 16 {k}^{2} + 16k - 24k - 16 {k}^{2} > 0$

$\\ \implies \: \sf 4 - 8k > 0$

$\\ \implies \: \sf 4 > 8k$

$\\ \implies \: \sf 8k < 4$

$\\ \implies \: \sf k < \dfrac{4}{8}$

$\\ \implies \: \sf k < \dfrac{1}{2}$

• Hence –

$\\ \implies \large{ \boxed{ \: \sf k \in \left( - \infty \:,\: \dfrac{1}{2} \right)}}$

$\\ \rule{220}{2}$

▪︎ If roots will be equal then –

$\\ \longrightarrow \: \large{ \boxed{ \sf D = 0 }}$

$\\ \longrightarrow \: \large{ \boxed{ \sf {b}^{2} - 4ac = 0 }}$

• So that –

$\\ \implies \: \sf 4-8k= 0$

$\\ \implies \: \sf k = \dfrac{4}{8}$

$\\ \implies \large{ \boxed{ \: \sf k = \dfrac{1}{2} }}$