find the value of K for which the roots of the equation 8k x(x-1)+1=0
Answers
Step-by-step explanation:
Givenarrow8kx(x−1)+1=0
\begin{gathered}\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}\end{gathered}
Now,
⟹8kx(x−1)+1=0
⟹8kx
2
−8kx+1=0
Here,
⟶Coefficientofx
2
(a)=8k
⟶Coefficientofx(b)=−8k
⟶Constantterm(c)=1
\begin{gathered}\textsf{To have equal zeroes :} \\ \\ \mathsf{\implies D \: = \: 0} \\ \\ \mathsf{\implies b^{2} \: - \: 4ac \: = \: 0} \\ \\ \textsf{Plug the value of a , b and c.}\end{gathered}
To have equal zeroes :
⟹D=0
⟹b
2
−4ac=0
Plug the value of a , b and c.
\begin{gathered}\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} }\end{gathered}
⟹(−8k)
2
−4×8k×1=0
⟹64k
2
−32k=0
⟹32k(2k−1)=0
⟹(2k−1)=
32k
0
⟹2k−1=0
⟹2k=1
∴k=
2
1
\underline{\textsf{Verification : }}
Verification :
\begin{gathered}\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}\end{gathered}
Plug the value of k in above equation :
⟹8kx(x−1)+1=0
⟹8(
2
1
)x(x−1)+1=0
⟹4x(x−1)+1=0
⟹4x
2
−4x+1=0
⟹(2x)
2
−2⋅2x⋅1+(1)
2
=0
\begin{gathered}\textsf{Using Algebraic Identity : } \\ \\ \boxed{\mathsf{\implies {a}^{2} \: - \: 2ab \: + \: {b}^{2} \: = \: (a \: - \: b)^{2}}}\end{gathered}
Using Algebraic Identity :
⟹a
2
−2ab+b
2
=(a−b)
2
\begin{gathered}\mathsf{\implies (2x \: - \: 1)^{2} \: = \: 0 } \\ \\ \mathsf{\implies (2x \: - \: 1)(2x \: - \: 1) \: = \: 0} \\ \\ \underline{\mathsf{By \: Zero \: Product \: Rule \: : }}\end{gathered}
⟹(2x−1)
2
=0
⟹(2x−1)(2x−1)=0
ByZeroProductRule:
\begin{gathered}\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} }\end{gathered}
⟹(2x−1)=0or⟹(2x−1)=0
⟹2x=1or⟹2x=1
⟹x=
2
1
or⟹x=
2
1
\textsf{Now, we got that zeroes are real and equal.}Now, we got that zeroes are real and equal.
\underline{\textsf{Verified !!}}
Verified !!
\boxed{\boxed{\mathsf{The \: Value \: of \: k \: is \: \dfrac{1}{2}.}}}
TheValueofkis
2
1
.
\begin{gathered}\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.}\end{gathered}
Note -: While finding the value of k , we
won’t take another value as 0.
If we do so then the equation
won’t remain a Quadratic Equation.