Solve 4x+9+7/x=29

$$4x+9+ \frac{ 7 }{ x } = 29$$

Solve for x

$$x = \frac{3 \sqrt{2} + 5}{2} \approx 4.621320344$$

$$x=\frac{5-3\sqrt{2}}{2}\approx 0.378679656$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$4xx+x\times 9+7=29x$$

Multiply x and x to get x^{2}.

$$4x^{2}+x\times 9+7=29x$$

Subtract 29x from both sides.

$$4x^{2}+x\times 9+7-29x=0$$

Combine x\times 9 and -29x to get -20x.

$$4x^{2}-20x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -20 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 7}}{2\times 4}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 7}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 7}}{2\times 4}$$

Multiply -16 times 7.

$$x=\frac{-\left(-20\right)±\sqrt{400-112}}{2\times 4}$$

Add 400 to -112.

$$x=\frac{-\left(-20\right)±\sqrt{288}}{2\times 4}$$

Take the square root of 288.

$$x=\frac{-\left(-20\right)±12\sqrt{2}}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±12\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±12\sqrt{2}}{8}$$

Now solve the equation x=\frac{20±12\sqrt{2}}{8} when ± is plus. Add 20 to 12\sqrt{2}.

$$x=\frac{12\sqrt{2}+20}{8}$$

Divide 20+12\sqrt{2} by 8.

$$x=\frac{3\sqrt{2}+5}{2}$$

Now solve the equation x=\frac{20±12\sqrt{2}}{8} when ± is minus. Subtract 12\sqrt{2} from 20.

$$x=\frac{20-12\sqrt{2}}{8}$$

Divide 20-12\sqrt{2} by 8.

$$x=\frac{5-3\sqrt{2}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{2}+5}{2} x=\frac{5-3\sqrt{2}}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$4xx+x\times 9+7=29x$$

Multiply x and x to get x^{2}.

$$4x^{2}+x\times 9+7=29x$$

Subtract 29x from both sides.

$$4x^{2}+x\times 9+7-29x=0$$

Combine x\times 9 and -29x to get -20x.

$$4x^{2}-20x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-20x=-7$$

Divide both sides by 4.

$$\frac{4x^{2}-20x}{4}=-\frac{7}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{20}{4}\right)x=-\frac{7}{4}$$

Divide -20 by 4.

$$x^{2}-5x=-\frac{7}{4}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{7}{4}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{-7+25}{4}$$

Add -\frac{7}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{2}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{2}}$$

Simplify.

$$x-\frac{5}{2}=\frac{3\sqrt{2}}{2} x-\frac{5}{2}=-\frac{3\sqrt{2}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{3\sqrt{2}+5}{2} x=\frac{5-3\sqrt{2}}{2}$$