Solve by Completing the Square 25x^2-10x=4

$25 x^{2} - 10 x = 4$

Step 1

Divide each term in $25 x^{2} - 10 x = 4$ by $25$ and simplify.

Divide each term in $25 x^{2} - 10 x = 4$ by $25$ .

$\frac{25 x^{2}}{25} + \frac{- 10 x}{25} = \frac{4}{25}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 x^{2}}{25} + \frac{- 10 x}{25} = \frac{4}{25}$

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{- 10 x}{25} = \frac{4}{25}$

Cancel the common factor of $- 10$ and $25$ .

Factor $5$ out of $- 10 x$ .

$x^{2} + \frac{5 (- 2 x)}{25} = \frac{4}{25}$

Cancel the common factors.

Factor $5$ out of $25$ .

$x^{2} + \frac{5 (- 2 x)}{5 (5)} = \frac{4}{25}$

Cancel the common factor.

$x^{2} + \frac{5 (- 2 x)}{5 \cdot 5} = \frac{4}{25}$

Rewrite the expression.

$x^{2} + \frac{- 2 x}{5} = \frac{4}{25}$

Move the negative in front of the fraction.

$x^{2} - \frac{2 x}{5} = \frac{4}{25}$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{1}{5})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - \frac{2 x}{5} + {(- \frac{1}{5})}^{2} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{5}$ .

$x^{2} - \frac{2 x}{5} + {(- 1)}^{2} {(\frac{1}{5})}^{2} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

Apply the product rule to $\frac{1}{5}$ .

$x^{2} - \frac{2 x}{5} + {(- 1)}^{2} \frac{1^{2}}{5^{2}} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{2 x}{5} + 1 \frac{1^{2}}{5^{2}} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

Multiply $\frac{1^{2}}{5^{2}}$ by $1$ .

$x^{2} - \frac{2 x}{5} + \frac{1^{2}}{5^{2}} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

One to any power is one.

$x^{2} - \frac{2 x}{5} + \frac{1}{5^{2}} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

Raise $5$ to the power of $2$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + {(- \frac{1}{5})}^{2}$

Simplify the right side.

Simplify $\frac{4}{25} + {(- \frac{1}{5})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{5}$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + {(- 1)}^{2} {(\frac{1}{5})}^{2}$

Apply the product rule to $\frac{1}{5}$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + {(- 1)}^{2} \frac{1^{2}}{5^{2}}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + 1 \frac{1^{2}}{5^{2}}$

Multiply $\frac{1^{2}}{5^{2}}$ by $1$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + \frac{1^{2}}{5^{2}}$

One to any power is one.

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + \frac{1}{5^{2}}$

Raise $5$ to the power of $2$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4}{25} + \frac{1}{25}$

Simplify terms.

Combine the numerators over the common denominator.

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{4 + 1}{25}$

Add $4$ and $1$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{5}{25}$

Cancel the common factor of $5$ and $25$ .

Factor $5$ out of $5$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{5 (1)}{25}$

Cancel the common factors.

Factor $5$ out of $25$ .

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{5 \cdot 1}{5 \cdot 5}$

Cancel the common factor.

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{5 \cdot 1}{5 \cdot 5}$

Rewrite the expression.

$x^{2} - \frac{2 x}{5} + \frac{1}{25} = \frac{1}{5}$

Step 5

Factor the perfect trinomial square into ${(x - \frac{1}{5})}^{2}$ .

${(x - \frac{1}{5})}^{2} = \frac{1}{5}$

Step 6

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - \frac{1}{5} = \pm \sqrt{\frac{1}{5}}$

Simplify $\pm \sqrt{\frac{1}{5}}$ .

Rewrite $\sqrt{\frac{1}{5}}$ as $\frac{\sqrt{1}}{\sqrt{5}}$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{1}}{\sqrt{5}}$

Any root of $1$ is $1$ .

$x - \frac{1}{5} = \pm \frac{1}{\sqrt{5}}$

Multiply $\frac{1}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$x - \frac{1}{5} = \pm \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{\sqrt{5} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{{\sqrt{5}}^{1} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Add $1$ and $1$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{{\sqrt{5}}^{2}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{5^{\frac{2}{2}}}$

Rewrite the expression.

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{5^{1}}$

Evaluate the exponent.

$x - \frac{1}{5} = \pm \frac{\sqrt{5}}{5}$

Add $\frac{1}{5}$ to both sides of the equation.

$x = \pm \frac{\sqrt{5}}{5} + \frac{1}{5}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{5}}{5} + \frac{1}{5}$

Decimal Form:

$x = 0.64721359 \dots, - 0.24721359 \dots$

Solve by Completing the Square 25x^2-10x=4

Related Questions

Support LCMGCF.com by disabling your adblocker.