Question

$\frac{1}{4} k = 3(- \frac{1}{4} k+3)k =$

​

Answer 1

Question

Find the value of 'k' in the equation ⇒ $\frac{1}{4}k = 3(\frac{-1}{4}k+3)k$

$\rule{300}{1}$

Answer

Let's solve your equation step-by-step.

$\frac{1}{4}k = 3(\frac{-1}{4}k+3)k$

Step 1: Simplify both sides of the equation.

⇒ $\frac{1}{4}k = \frac{-3}{4}k^{2}+9k$

Step 2: Subtract $\frac{-3}{4} k^{2}+9k$ from both sides.

⇒ $\frac{1}{4}k - (\frac{-3}{4} k^{2}+9k)= \frac{-3}{4}k^{2}+9k-(\frac{-3}{4} k^{2}+9k)$

⇒ $\frac{3}{4}k^{2}+\frac{-35}{4}k=0$

Step 3: Factor left side of equation.

⇒ $k(\frac{3}{4}k+\frac{-35}{4})=0$

Step 4: Set factors equal to 0.

⇒ $k=0\ or \ \frac{3}{4}k +\frac{-35}{4} =0$

⇒ $k=0 \ or \ k=\frac{35}{3}$

Let's verify step-by-step.

First instance where k=0

⇒ $\frac{1}{4}k = 3(\frac{-1}{4}k+3)k$

⇒ $\frac{1}{4}\times 0 = 3(\frac{-1}{4}\times 0 +3)\times 0$

⇒ $0=3(0)(0)$

⇒ $0=0$

Second instance where k = ³⁵/₃

⇒ $\frac{1}{4}k = 3(\frac{-1}{4}k+3)k$

⇒ $\frac{1}{4}\times \frac{35}{3} = 3(\frac{-1}{4}\times \frac{35}{3} +3)\times \frac{35}{3}$

⇒ $\frac{35}{12} = 3(\frac{-35}{12} +3)\times \frac{35}{3}$

⇒ $\frac{35}{12} = 3(\frac{-35}{12} +\frac{36}{12} )\times \frac{35}{3}$

⇒ $\frac{35}{12} = 3\times \frac{1}{12} \times \frac{35}{3}$

⇒ $\frac{35}{12} = \frac{105}{36}$

⇒ $\frac{35}{12} = \frac{35}{12}$

∴LHS=RHS is both the cases.

$\bf \therefore k=0 \ or \ k=\frac{35}{3}\ in \ the \ equation =>\ \frac{1}{4}k = 3(\frac{-1}{4}k+3)k$

$\rule{300}{1}$

Answer 2

Step-by-step explanation:

Question

Find the value of 'k' in the equation ⇒ \frac{1}{4}k = 3(\frac{-1}{4}k+3)k

4

1

k=3(

4

−1

k+3)k

\rule{300}{1}

Answer

Let's solve your equation step-by-step.

\frac{1}{4}k = 3(\frac{-1}{4}k+3)k

4

1

k=3(

4

−1

k+3)k

Step 1: Simplify both sides of the equation.

⇒ \frac{1}{4}k = \frac{-3}{4}k^{2}+9k

4

1

k=

4

−3

k

2

+9k

Step 2: Subtract \frac{-3}{4} k^{2}+9k

4

−3

k

2

+9k from both sides.

⇒ \frac{1}{4}k - (\frac{-3}{4} k^{2}+9k)= \frac{-3}{4}k^{2}+9k-(\frac{-3}{4} k^{2}+9k)

4

1

k−(

4

−3

k

2

+9k)=

4

−3

k

2

+9k−(

4

−3

k

2

+9k)

⇒ \frac{3}{4}k^{2}+\frac{-35}{4}k=0

4

3

k

2

+

4

−35

k=0

Step 3: Factor left side of equation.

⇒ k(\frac{3}{4}k+\frac{-35}{4})=0k(

4

3

k+

4

−35

)=0

Step 4: Set factors equal to 0.

⇒ k=0\ or \ \frac{3}{4}k +\frac{-35}{4} =0k=0 or

4

3

k+

4

−35

=0

⇒ k=0 \ or \ k=\frac{35}{3}k=0 or k=

3

35

Let's verify step-by-step.

First instance where k=0

⇒ \frac{1}{4}k = 3(\frac{-1}{4}k+3)k

4

1

k=3(

4

−1

k+3)k

⇒ \frac{1}{4}\times 0 = 3(\frac{-1}{4}\times 0 +3)\times 0

4

1

×0=3(

4

−1

×0+3)×0

⇒ 0=3(0)(0)0=3(0)(0)

⇒ 0=00=0

Second instance where k = ³⁵/₃

⇒ \frac{1}{4}k = 3(\frac{-1}{4}k+3)k

4

1

k=3(

4

−1

k+3)k

⇒ \frac{1}{4}\times \frac{35}{3} = 3(\frac{-1}{4}\times \frac{35}{3} +3)\times \frac{35}{3}

4

1

×

3

35

=3(

4

−1

×

3

35

+3)×

3

35

⇒ \frac{35}{12} = 3(\frac{-35}{12} +3)\times \frac{35}{3}

12

35

=3(

12

−35

+3)×

3

35

⇒ \frac{35}{12} = 3(\frac{-35}{12} +\frac{36}{12} )\times \frac{35}{3}

12

35

=3(

12

−35

+

12

36

)×

3

35

⇒ \frac{35}{12} = 3\times \frac{1}{12} \times \frac{35}{3}

12

35

=3×

12

1

×

3

35

⇒ \frac{35}{12} = \frac{105}{36}

12

35

=

36

105

⇒ \frac{35}{12} = \frac{35}{12}

12

35

=

12

35

∴LHS=RHS is both the cases.

\bf \therefore k=0 \ or \ k=\frac{35}{3}\ in \ the \ equation = > \ \frac{1}{4}k = 3(\frac{-1}{4}k+3)k∴k=0 or k=

3

35

in the equation=>

4

1

k=3(

4

−1

k+3)k

\rule{300}{1}