Solve by Completing the Square 5x^2+14x=3

$5 x^{2} + 14 x = 3$

Step 1

Divide each term in $5 x^{2} + 14 x = 3$ by $5$ and simplify.

Divide each term in $5 x^{2} + 14 x = 3$ by $5$ .

$\frac{5 x^{2}}{5} + \frac{14 x}{5} = \frac{3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x^{2}}{5} + \frac{14 x}{5} = \frac{3}{5}$

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

$x^{2} + \frac{14 x}{5} = \frac{3}{5}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{7}{5})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + \frac{14 x}{5} + {(\frac{7}{5})}^{2} = \frac{3}{5} + {(\frac{7}{5})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{7}{5}$ .

$x^{2} + \frac{14 x}{5} + \frac{7^{2}}{5^{2}} = \frac{3}{5} + {(\frac{7}{5})}^{2}$

Raise $7$ to the power of $2$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{5^{2}} = \frac{3}{5} + {(\frac{7}{5})}^{2}$

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

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3}{5} + {(\frac{7}{5})}^{2}$

Simplify the right side.

Simplify $\frac{3}{5} + {(\frac{7}{5})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{7}{5}$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3}{5} + \frac{7^{2}}{5^{2}}$

Raise $7$ to the power of $2$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3}{5} + \frac{49}{5^{2}}$

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

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3}{5} + \frac{49}{25}$

To write $\frac{3}{5}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3}{5} \cdot \frac{5}{5} + \frac{49}{25}$

Write each expression with a common denominator of $25$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{5}$ by $\frac{5}{5}$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3 \cdot 5}{5 \cdot 5} + \frac{49}{25}$

Multiply $5$ by $5$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3 \cdot 5}{25} + \frac{49}{25}$

Combine the numerators over the common denominator.

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{3 \cdot 5 + 49}{25}$

Simplify the numerator.

Multiply $3$ by $5$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{15 + 49}{25}$

Add $15$ and $49$ .

$x^{2} + \frac{14 x}{5} + \frac{49}{25} = \frac{64}{25}$

Step 5

Factor the perfect trinomial square into ${(x + \frac{7}{5})}^{2}$ .

${(x + \frac{7}{5})}^{2} = \frac{64}{25}$

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.

$x + \frac{7}{5} = \pm \sqrt{\frac{64}{25}}$

Simplify $\pm \sqrt{\frac{64}{25}}$ .

Rewrite $\sqrt{\frac{64}{25}}$ as $\frac{\sqrt{64}}{\sqrt{25}}$ .

$x + \frac{7}{5} = \pm \frac{\sqrt{64}}{\sqrt{25}}$

Simplify the numerator.

Rewrite $64$ as $8^{2}$ .

$x + \frac{7}{5} = \pm \frac{\sqrt{8^{2}}}{\sqrt{25}}$

Pull terms out from under the radical, assuming positive real numbers.

$x + \frac{7}{5} = \pm \frac{8}{\sqrt{25}}$

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

$x + \frac{7}{5} = \pm \frac{8}{\sqrt{5^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$x + \frac{7}{5} = \pm \frac{8}{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x + \frac{7}{5} = \frac{8}{5}$

Move all terms not containing $x$ to the right side of the equation.

Subtract $\frac{7}{5}$ from both sides of the equation.

$x = \frac{8}{5} - \frac{7}{5}$

Combine the numerators over the common denominator.

$x = \frac{8 - 7}{5}$

Subtract $7$ from $8$ .

$x = \frac{1}{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$x + \frac{7}{5} = - \frac{8}{5}$

Move all terms not containing $x$ to the right side of the equation.

Subtract $\frac{7}{5}$ from both sides of the equation.

$x = - \frac{8}{5} - \frac{7}{5}$

Combine the numerators over the common denominator.

$x = \frac{- 8 - 7}{5}$

Subtract $7$ from $- 8$ .

$x = \frac{- 15}{5}$

Divide $- 15$ by $5$ .

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{1}{5}, - 3$

Solve by Completing the Square 5x^2+14x=3

Related Questions

Support LCMGCF.com by disabling your adblocker.