Solve 4/x+15=35/x^2

$$\frac { 4 } { x } + 15 = \frac { 35 } { x ^ { 2 } }$$

Solve for x

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x,x^{2}.

$$x\times 4+x^{2}\times 15=35$$

Subtract 35 from both sides.

$$x\times 4+x^{2}\times 15-35=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$15x^{2}+4x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=4 ab=15\left(-35\right)=-525$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -525.

$$-1,525 -3,175 -5,105 -7,75 -15,35 -21,25$$

Calculate the sum for each pair.

$$-1+525=524 -3+175=172 -5+105=100 -7+75=68 -15+35=20 -21+25=4$$

The solution is the pair that gives sum 4.

$$a=-21 b=25$$

Rewrite 15x^{2}+4x-35 as \left(15x^{2}-21x\right)+\left(25x-35\right).

$$\left(15x^{2}-21x\right)+\left(25x-35\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(5x-7\right)+5\left(5x-7\right)$$

Factor out common term 5x-7 by using distributive property.

$$\left(5x-7\right)\left(3x+5\right)$$

To find equation solutions, solve 5x-7=0 and 3x+5=0.

$$x=\frac{7}{5} x=-\frac{5}{3}$$

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^{2}, the least common multiple of x,x^{2}.

$$x\times 4+x^{2}\times 15=35$$

Subtract 35 from both sides.

$$x\times 4+x^{2}\times 15-35=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$15x^{2}+4x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 4 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 15\left(-35\right)}}{2\times 15}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 15\left(-35\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-4±\sqrt{16-60\left(-35\right)}}{2\times 15}$$

Multiply -60 times -35.

$$x=\frac{-4±\sqrt{16+2100}}{2\times 15}$$

Add 16 to 2100.

$$x=\frac{-4±\sqrt{2116}}{2\times 15}$$

Take the square root of 2116.

$$x=\frac{-4±46}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-4±46}{30}$$

Now solve the equation x=\frac{-4±46}{30} when ± is plus. Add -4 to 46.

$$x=\frac{42}{30}$$

Reduce the fraction \frac{42}{30} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-4±46}{30} when ± is minus. Subtract 46 from -4.

$$x=-\frac{50}{30}$$

Reduce the fraction \frac{-50}{30} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$x=\frac{7}{5} x=-\frac{5}{3}$$

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^{2}, the least common multiple of x,x^{2}.

$$x\times 4+x^{2}\times 15=35$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$15x^{2}+4x=35$$

Divide both sides by 15.

$$\frac{15x^{2}+4x}{15}=\frac{35}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\frac{4}{15}x=\frac{35}{15}$$

Reduce the fraction \frac{35}{15} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{4}{15}x=\frac{7}{3}$$

Divide \frac{4}{15}, the coefficient of the x term, by 2 to get \frac{2}{15}. Then add the square of \frac{2}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{15}x+\left(\frac{2}{15}\right)^{2}=\frac{7}{3}+\left(\frac{2}{15}\right)^{2}$$

Square \frac{2}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{15}x+\frac{4}{225}=\frac{7}{3}+\frac{4}{225}$$

Add \frac{7}{3} to \frac{4}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{15}x+\frac{4}{225}=\frac{529}{225}$$

Factor x^{2}+\frac{4}{15}x+\frac{4}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{15}\right)^{2}=\frac{529}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{15}\right)^{2}}=\sqrt{\frac{529}{225}}$$

Simplify.

$$x+\frac{2}{15}=\frac{23}{15} x+\frac{2}{15}=-\frac{23}{15}$$

Subtract \frac{2}{15} from both sides of the equation.

$$x=\frac{7}{5} x=-\frac{5}{3}$$