Question

The quadratic equation 2kx^ 2 +4x+k-1=0, has real and equal roots, find k.​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ 2kx² + 4x + k - 1 = 0.

Real and equal roots.

As we know that,

D = Discriminant Or b² - 4ac.

For real and equal roots : D = 0.

⇒ (4)² - 4(2k)(k - 1) = 0.

⇒ 16 - 8k(k - 1) = 0.

⇒ 16 - 8k² + 8k = 0.

⇒ - 8k² + 8k + 16 = 0.

⇒ 8k² - 8k - 16 = 0.

⇒ k² - k - 2 = 0.

⇒ k² - 2k + k - 2 = 0.

⇒ k(k - 2) + 1(k - 2) = 0.

⇒ (k + 1)(k - 2) = 0.

⇒ k = - 1   and   k = 2.

                                                                                                                       

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

$\pink{ \boxed{\boxed{\begin{array}{cc} \maltese \bf \: \: given \\ \bf \: the \: quadratic \: euation \: : \: \\ \\ \bf \: 2k {x}^{2} + 4x + k- 1 = 0 \\ \\ \blue{\sf \rightarrow \:we \: have \: to \: find \: the \: value \: of \: \: k }\\ \\ \bf \: when \: roots \: of \:the\:equation \: \: are \\ \orange{ \bf \: real \: \: and \: equal} \end{array}}}} \\ \\ \\ {\pink{ \boxed{\boxed{\begin{array}{cc} \maltese \bf \: Formula : \\ \\ \bf if \: a {x}^{2} + bx + c = 0 \: \: i s\: a \: qudratic \\ \bf \: equation \: and \: its \: roots \: are \: \\ \orange{ \bf \: real \: and \: equal} \\ \bf \: then \\ \\ \blue{\boxed{ \begin{array}{cc}\bf \: discriminant \: (D )= 0 \\ \\ \sf \: and \\ \\ \bf \: discriminant \: (D) = {b}^{2} - 4ac \end{array}}} \end{array}}}}}\\ \\ \\ \blue{ \boxed{\boxed{\begin{array}{cc} \maltese \sf \:according \: to \: the \: question \\ \\ \bf \: b = 4 \\ \bf \: a = 2k \\ \bf \: c = k - 1 \\ \\ \sf \: for \: real \: and \: equal \: roots \\ \\ \bf {b}^{2} - 4ac = 0 \\ \\ \bf \implies \: {(4)}^{2} - 4 \times 2k \times (k - 1) = 0 \\ \\ \bf \implies \: 16 - 8k(k - 1) = 0 \\ \\ \bf \implies \: 16 - 8 {k}^{2} + 8k = 0 \\ \\ \bf \implies \: 8 {k}^{2} - 8k - 16 = 0 \\ \\ \bf \implies \: {k}^{2} - k - 2 = 0 \\ \\ \bf \implies \: {k}^{2} - 2k + k - 2 = 0 \\ \\\bf \implies \: k(k - 2) + 1(k - 2) = 0 \\ \\ \bf \implies \:( k - 2)(k + 1) = 0 \\ \\ \pink{ \boxed{\boxed{\begin{array}{c | c} \bf \: either \:& \bf \: or \: \\ \bf \: k - 2 = 0& \bf \: k + 1 = 0 \\ \bf \therefore \:k = 2 & \bf \therefore \: k = - 1 \:\end{array}}}} \\ \\ \bf \therefore \:k = 2 \: \: or \: \: - 1\end{array}}}}$