Solve 6x^2-35x-35/x+6/x^2+62=0

$$6 x ^ { 2 } - 35 x - \frac { 35 } { x } + \frac { 6 } { x ^ { 2 } } + 62 = 0$$

Solve for x

$$x=2$$

$$x=3$$

$$x=\frac{1}{2}=0.5$$

$$x=\frac{1}{3}\approx 0.333333333$$

Solution Steps

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}.

$$6x^{2}x^{2}-35xx^{2}-x\times 35+6+x^{2}\times 62=0$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$6x^{4}-35xx^{2}-x\times 35+6+x^{2}\times 62=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$6x^{4}-35x^{3}-x\times 35+6+x^{2}\times 62=0$$

Multiply -1 and 35 to get -35.

$$6x^{4}-35x^{3}-35x+6+x^{2}\times 62=0$$

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

$$6x^{4}-35x^{3}+62x^{2}-35x+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±1,±2,±3,±6,±\frac{1}{2},±\frac{3}{2},±\frac{1}{3},±\frac{2}{3},±\frac{1}{6}$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{4}-35x^{3}+62x^{2}-35x+6 by x-2 to get 6x^{3}-23x^{2}+16x-3. Solve the equation where the result equals to 0.

$$6x^{3}-23x^{2}+16x-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1,±\frac{3}{2},±3,±\frac{1}{6},±\frac{1}{3}$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{3}-23x^{2}+16x-3 by x-3 to get 6x^{2}-5x+1. Solve the equation where the result equals to 0.

$$6x^{2}-5x+1=0$$

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 6 for a, -5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\times 1}}{2\times 6}$$

Do the calculations.

$$x=\frac{5±1}{12}$$

Solve the equation 6x^{2}-5x+1=0 when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=\frac{1}{2}$$

List all found solutions.

$$x=2 x=3 x=\frac{1}{3} x=\frac{1}{2}$$