Question

Find the value of k for which the roots of the equation 8kx ( x - 1 ) + 1 = 0 are real and equal.

Class 10

Quadratic Equation

Answer 1

8kx( x - 1 ) + 1 = 0

8kx² - 8kx + 1 = 0

If the roots are real and equal

⇒ the discriminant = 0

Solve k

b² - 4ac = 0

(-8k)² - 4(8k)(1) =  0

64k² - 32k = 0

32k (2k - 1) = 0

k = 0 or 2k - 1 = 0

k = 0 or k = 1/2

Answer: k = 0 or k = 1/2

Answer 2

$\underline{\underline{\large{\mathfrak{Solution : }}}}$




$\underline{\mathsf{Given \rightarrow 8kx(x \: - \: 1 ) \: + \: 1 \: = \: 0}}$




$\textsf{Now,} \\ \\ \mathsf{\implies 8kx (x \: - \: 1) \: + \: 1 \: = \: 0} \\ \\ \mathsf{\implies 8k{x}^{2} \: - \: 8kx \: + \: 1 \: = \: 0} \\ \\ \mathsf{Here,} \\ \\ \mathsf{\longrightarrow Coefficient \: of \: {x}^{2}(a) \: = \: 8k} \\ \\ \mathsf{\longrightarrow Coefficient \: of \: {x}(b) \: = \: -8k} \\ \\ \mathsf{\longrightarrow Constant \: term(c) \: = \: 1}$




$\textsf{To have equal zeroes :} \\ \\ \mathsf{\implies D \: = \: 0} \\ \\ \mathsf{\implies b^{2} \: - \: 4ac \: = \: 0} \\ \\ \textsf{Plug the value of a , b and c.}$




$\mathsf{\implies (-8k)^{2} \: - \: 4 \: \times \: 8k \: \times \: 1 \: = \: 0} \\ \\ \mathsf{ \implies 64 {k}^{2} \: - \: 32k \: = \: 0 \: } \\ \\ \mathsf{ \implies32k(2k \: - \: 1) \: = \: 0} \\ \\ \mathsf{ \implies(2k \: - \: 1) \: = \: \dfrac{0}{32k} } \\ \\ \mathsf{ \implies2k \: - \: 1 \: = \: 0} \\ \\ \mathsf{ \implies2k \: = \: 1} \\ \\ \mathsf{ \therefore \quad {k} \: = \: \dfrac{1}{2} }$




$\underline{\textsf{Verification : }}$




$\textsf{Plug the value of k in above equation : } \\ \\ \mathsf{\implies 8kx(x \: - \: 1) \: + \: 1 \: = \: 0} \\ \\ \mathsf{\implies 8\Bigg( \dfrac{1}{2} \Bigg) x ( x \: - \: 1) \: + \: 1 \: = \: 0} \\ \\ \mathsf{\implies 4x(x \: - \: 1 ) \: + \: 1 \: = \: 0} \\ \\ \mathsf{\implies 4{x}^{2} \: - \: 4x \: + \: 1 \: = \: 0} \\ \\ \mathsf{\implies (2x)^{2} \: - \: 2 \: \cdot \: 2x \: \cdot \: 1 \: + \: {(1)}^{2} \: = \: 0}$




$\textsf{Using Algebraic Identity : } \\ \\ \boxed{\mathsf{\implies {a}^{2} \: - \: 2ab \: + \: {b}^{2} \: = \: (a \: - \: b)^{2}}}$




$\mathsf{\implies (2x \: - \: 1)^{2} \: = \: 0 } \\ \\ \mathsf{\implies (2x \: - \: 1)(2x \: - \: 1) \: = \: 0} \\ \\ \underline{\mathsf{By \: Zero \: Product \: Rule \: : }}$




$\mathsf{\implies (2x \: - \: 1) \: = \: 0 \quad \: or \: \implies (2x \: - \: 1) \: = \: 0} \\ \\ \mathsf{ \implies 2x \: = \: 1 \quad \: or \: \implies 2x \: = \: 1} \\ \\ \mathsf{ \implies x \: = \: \dfrac{1}{2} \quad \: or \: \implies x \: = \: \dfrac{1}{2} }$




$\textsf{Now, we got that zeroes are real and equal.}$




$\underline{\textsf{Verified !!}}$




$\boxed{\boxed{\mathsf{The \: Value \: of \: k \: is \: \dfrac{1}{2}.}}}$




$\textsf{Note -: While finding the value of k , we} \\ \textsf{ won't take another value as 0.}\\\textsf{If we do so then the equation} \\ \textsf{ won't remain a Quadratic Equation.}$