Solve sqrt{5x+2}-4/sqrt{5x+2}=5

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

Solve for x

$$x=\frac{\sqrt{41}}{2}+\frac{29}{10}\approx 6.101562119$$

Solution Steps

Subtract -\frac{4}{\sqrt{5x+2}} from both sides of the equation.

$$\sqrt{5x+2}=5-\left(-\frac{4}{\sqrt{5x+2}}\right)$$

Multiply -1 and -1 to get 1.

$$\sqrt{5x+2}=5+\frac{4}{\sqrt{5x+2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{\sqrt{5x+2}}{\sqrt{5x+2}}.

$$\sqrt{5x+2}=\frac{5\sqrt{5x+2}}{\sqrt{5x+2}}+\frac{4}{\sqrt{5x+2}}$$

Since \frac{5\sqrt{5x+2}}{\sqrt{5x+2}} and \frac{4}{\sqrt{5x+2}} have the same denominator, add them by adding their numerators.

$$\sqrt{5x+2}=\frac{5\sqrt{5x+2}+4}{\sqrt{5x+2}}$$

Square both sides of the equation.

$$\left(\sqrt{5x+2}\right)^{2}=\left(\frac{5\sqrt{5x+2}+4}{\sqrt{5x+2}}\right)^{2}$$

Calculate \sqrt{5x+2} to the power of 2 and get 5x+2.

$$5x+2=\left(\frac{5\sqrt{5x+2}+4}{\sqrt{5x+2}}\right)^{2}$$

To raise \frac{5\sqrt{5x+2}+4}{\sqrt{5x+2}} to a power, raise both numerator and denominator to the power and then divide.

$$5x+2=\frac{\left(5\sqrt{5x+2}+4\right)^{2}}{\left(\sqrt{5x+2}\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5\sqrt{5x+2}+4\right)^{2}.

$$5x+2=\frac{25\left(\sqrt{5x+2}\right)^{2}+40\sqrt{5x+2}+16}{\left(\sqrt{5x+2}\right)^{2}}$$

Calculate \sqrt{5x+2} to the power of 2 and get 5x+2.

$$5x+2=\frac{25\left(5x+2\right)+40\sqrt{5x+2}+16}{\left(\sqrt{5x+2}\right)^{2}}$$

Use the distributive property to multiply 25 by 5x+2.

$$5x+2=\frac{125x+50+40\sqrt{5x+2}+16}{\left(\sqrt{5x+2}\right)^{2}}$$

Add 50 and 16 to get 66.

$$5x+2=\frac{125x+66+40\sqrt{5x+2}}{\left(\sqrt{5x+2}\right)^{2}}$$

Calculate \sqrt{5x+2} to the power of 2 and get 5x+2.

$$5x+2=\frac{125x+66+40\sqrt{5x+2}}{5x+2}$$

Multiply both sides of the equation by 5x+2.

$$5x\left(5x+2\right)+\left(5x+2\right)\times 2=125x+66+40\sqrt{5x+2}$$

Subtract 125x+66 from both sides of the equation.

$$5x\left(5x+2\right)+\left(5x+2\right)\times 2-\left(125x+66\right)=40\sqrt{5x+2}$$

Use the distributive property to multiply 5x by 5x+2.

$$25x^{2}+10x+\left(5x+2\right)\times 2-\left(125x+66\right)=40\sqrt{5x+2}$$

Use the distributive property to multiply 5x+2 by 2.

$$25x^{2}+10x+10x+4-\left(125x+66\right)=40\sqrt{5x+2}$$

Combine 10x and 10x to get 20x.

$$25x^{2}+20x+4-\left(125x+66\right)=40\sqrt{5x+2}$$

To find the opposite of 125x+66, find the opposite of each term.

$$25x^{2}+20x+4-125x-66=40\sqrt{5x+2}$$

Combine 20x and -125x to get -105x.

$$25x^{2}-105x+4-66=40\sqrt{5x+2}$$

Subtract 66 from 4 to get -62.

$$25x^{2}-105x-62=40\sqrt{5x+2}$$

Square both sides of the equation.

$$\left(25x^{2}-105x-62\right)^{2}=\left(40\sqrt{5x+2}\right)^{2}$$

Square 25x^{2}-105x-62.

$$625x^{4}-5250x^{3}+7925x^{2}+13020x+3844=\left(40\sqrt{5x+2}\right)^{2}$$

Expand \left(40\sqrt{5x+2}\right)^{2}.

$$625x^{4}-5250x^{3}+7925x^{2}+13020x+3844=40^{2}\left(\sqrt{5x+2}\right)^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$625x^{4}-5250x^{3}+7925x^{2}+13020x+3844=1600\left(\sqrt{5x+2}\right)^{2}$$

Calculate \sqrt{5x+2} to the power of 2 and get 5x+2.

$$625x^{4}-5250x^{3}+7925x^{2}+13020x+3844=1600\left(5x+2\right)$$

Use the distributive property to multiply 1600 by 5x+2.

$$625x^{4}-5250x^{3}+7925x^{2}+13020x+3844=8000x+3200$$

Subtract 8000x from both sides.

$$625x^{4}-5250x^{3}+7925x^{2}+13020x+3844-8000x=3200$$

Combine 13020x and -8000x to get 5020x.

$$625x^{4}-5250x^{3}+7925x^{2}+5020x+3844=3200$$

Subtract 3200 from both sides.

$$625x^{4}-5250x^{3}+7925x^{2}+5020x+3844-3200=0$$

Subtract 3200 from 3844 to get 644.

$$625x^{4}-5250x^{3}+7925x^{2}+5020x+644=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 644 and q divides the leading coefficient 625. List all candidates \frac{p}{q}.

$$±\frac{644}{625},±\frac{644}{125},±\frac{644}{25},±\frac{644}{5},±644,±\frac{322}{625},±\frac{322}{125},±\frac{322}{25},±\frac{322}{5},±322,±\frac{161}{625},±\frac{161}{125},±\frac{161}{25},±\frac{161}{5},±161,±\frac{92}{625},±\frac{92}{125},±\frac{92}{25},±\frac{92}{5},±92,±\frac{46}{625},±\frac{46}{125},±\frac{46}{25},±\frac{46}{5},±46,±\frac{28}{625},±\frac{28}{125},±\frac{28}{25},±\frac{28}{5},±28,±\frac{23}{625},±\frac{23}{125},±\frac{23}{25},±\frac{23}{5},±23,±\frac{14}{625},±\frac{14}{125},±\frac{14}{25},±\frac{14}{5},±14,±\frac{7}{625},±\frac{7}{125},±\frac{7}{25},±\frac{7}{5},±7,±\frac{4}{625},±\frac{4}{125},±\frac{4}{25},±\frac{4}{5},±4,±\frac{2}{625},±\frac{2}{125},±\frac{2}{25},±\frac{2}{5},±2,±\frac{1}{625},±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 625x^{4}-5250x^{3}+7925x^{2}+5020x+644 by 5\left(x+\frac{1}{5}\right)=5x+1 to get 125x^{3}-1075x^{2}+1800x+644. Solve the equation where the result equals to 0.

$$125x^{3}-1075x^{2}+1800x+644=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 644 and q divides the leading coefficient 125. List all candidates \frac{p}{q}.

$$±\frac{644}{125},±\frac{644}{25},±\frac{644}{5},±644,±\frac{322}{125},±\frac{322}{25},±\frac{322}{5},±322,±\frac{161}{125},±\frac{161}{25},±\frac{161}{5},±161,±\frac{92}{125},±\frac{92}{25},±\frac{92}{5},±92,±\frac{46}{125},±\frac{46}{25},±\frac{46}{5},±46,±\frac{28}{125},±\frac{28}{25},±\frac{28}{5},±28,±\frac{23}{125},±\frac{23}{25},±\frac{23}{5},±23,±\frac{14}{125},±\frac{14}{25},±\frac{14}{5},±14,±\frac{7}{125},±\frac{7}{25},±\frac{7}{5},±7,±\frac{4}{125},±\frac{4}{25},±\frac{4}{5},±4,±\frac{2}{125},±\frac{2}{25},±\frac{2}{5},±2,±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{14}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 125x^{3}-1075x^{2}+1800x+644 by 5\left(x-\frac{14}{5}\right)=5x-14 to get 25x^{2}-145x-46. Solve the equation where the result equals to 0.

$$25x^{2}-145x-46=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 25 for a, -145 for b, and -46 for c in the quadratic formula.

$$x=\frac{-\left(-145\right)±\sqrt{\left(-145\right)^{2}-4\times 25\left(-46\right)}}{2\times 25}$$

Do the calculations.

$$x=\frac{145±25\sqrt{41}}{50}$$

Solve the equation 25x^{2}-145x-46=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{41}}{2}+\frac{29}{10} x=\frac{\sqrt{41}}{2}+\frac{29}{10}$$

List all found solutions.

$$x=-\frac{1}{5} x=\frac{14}{5} x=-\frac{\sqrt{41}}{2}+\frac{29}{10} x=\frac{\sqrt{41}}{2}+\frac{29}{10}$$

Substitute -\frac{1}{5} for x in the equation \sqrt{5x+2}-\frac{4}{\sqrt{5x+2}}=5.

$$\sqrt{5\left(-\frac{1}{5}\right)+2}-\frac{4}{\sqrt{5\left(-\frac{1}{5}\right)+2}}=5$$

Simplify. The value x=-\frac{1}{5} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=5$$

Substitute \frac{14}{5} for x in the equation \sqrt{5x+2}-\frac{4}{\sqrt{5x+2}}=5.

$$\sqrt{5\times \frac{14}{5}+2}-\frac{4}{\sqrt{5\times \frac{14}{5}+2}}=5$$

Simplify. The value x=\frac{14}{5} does not satisfy the equation.

$$3=5$$

Substitute -\frac{\sqrt{41}}{2}+\frac{29}{10} for x in the equation \sqrt{5x+2}-\frac{4}{\sqrt{5x+2}}=5.

$$\sqrt{5\left(-\frac{\sqrt{41}}{2}+\frac{29}{10}\right)+2}-\frac{4}{\sqrt{5\left(-\frac{\sqrt{41}}{2}+\frac{29}{10}\right)+2}}=5$$

Simplify. The value x=-\frac{\sqrt{41}}{2}+\frac{29}{10} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-5=5$$

Substitute \frac{\sqrt{41}}{2}+\frac{29}{10} for x in the equation \sqrt{5x+2}-\frac{4}{\sqrt{5x+2}}=5.

$$\sqrt{5\left(\frac{\sqrt{41}}{2}+\frac{29}{10}\right)+2}-\frac{4}{\sqrt{5\left(\frac{\sqrt{41}}{2}+\frac{29}{10}\right)+2}}=5$$

Simplify. The value x=\frac{\sqrt{41}}{2}+\frac{29}{10} satisfies the equation.

$$5=5$$

Equation \sqrt{5x+2}=\frac{5\sqrt{5x+2}+4}{\sqrt{5x+2}} has a unique solution.

$$x=\frac{\sqrt{41}}{2}+\frac{29}{10}$$