Solve by Completing the Square x^2+11x-5=0

$x^{2} + 11 x - 5 = 0$

Step 1

Add $5$ to both sides of the equation.

$x^{2} + 11 x = 5$

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{11}{2})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 11 x + {(\frac{11}{2})}^{2} = 5 + {(\frac{11}{2})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{11}{2}$ .

$x^{2} + 11 x + \frac{11^{2}}{2^{2}} = 5 + {(\frac{11}{2})}^{2}$

Raise $11$ to the power of $2$ .

$x^{2} + 11 x + \frac{121}{2^{2}} = 5 + {(\frac{11}{2})}^{2}$

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

$x^{2} + 11 x + \frac{121}{4} = 5 + {(\frac{11}{2})}^{2}$

Simplify the right side.

Simplify $5 + {(\frac{11}{2})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{11}{2}$ .

$x^{2} + 11 x + \frac{121}{4} = 5 + \frac{11^{2}}{2^{2}}$

Raise $11$ to the power of $2$ .

$x^{2} + 11 x + \frac{121}{4} = 5 + \frac{121}{2^{2}}$

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

$x^{2} + 11 x + \frac{121}{4} = 5 + \frac{121}{4}$

To write $5$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} + 11 x + \frac{121}{4} = 5 \cdot \frac{4}{4} + \frac{121}{4}$

Combine $5$ and $\frac{4}{4}$ .

$x^{2} + 11 x + \frac{121}{4} = \frac{5 \cdot 4}{4} + \frac{121}{4}$

Combine the numerators over the common denominator.

$x^{2} + 11 x + \frac{121}{4} = \frac{5 \cdot 4 + 121}{4}$

Simplify the numerator.

Multiply $5$ by $4$ .

$x^{2} + 11 x + \frac{121}{4} = \frac{20 + 121}{4}$

Add $20$ and $121$ .

$x^{2} + 11 x + \frac{121}{4} = \frac{141}{4}$

Step 5

Factor the perfect trinomial square into ${(x + \frac{11}{2})}^{2}$ .

${(x + \frac{11}{2})}^{2} = \frac{141}{4}$

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.

$x + \frac{11}{2} = \pm \sqrt{\frac{141}{4}}$

Simplify $\pm \sqrt{\frac{141}{4}}$ .

Rewrite $\sqrt{\frac{141}{4}}$ as $\frac{\sqrt{141}}{\sqrt{4}}$ .

$x + \frac{11}{2} = \pm \frac{\sqrt{141}}{\sqrt{4}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$x + \frac{11}{2} = \pm \frac{\sqrt{141}}{\sqrt{2^{2}}}$

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

$x + \frac{11}{2} = \pm \frac{\sqrt{141}}{2}$

Subtract $\frac{11}{2}$ from both sides of the equation.

$x = \pm \frac{\sqrt{141}}{2} - \frac{11}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{141}}{2} - \frac{11}{2}$

Decimal Form:

$x = 0.43717104 \dots, - 11.43717104 \dots$

Solve by Completing the Square x^2+11x-5=0

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