Solve 3=(2x^2+1)x

$$3 = ( 2 x ^ { 2 } + 1 ) x$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{5}i-1}{2}\approx -0.5-1.118033989i$$

$$x=\frac{-1+\sqrt{5}i}{2}\approx -0.5+1.118033989i$$

$$x=1$$

Solution Steps

Use the distributive property to multiply 2x^{2}+1 by x.

$$3=2x^{3}+x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{3}+x=3$$

Subtract 3 from both sides.

$$2x^{3}+x-3=0$$

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 2. List all candidates \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=1$$

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

$$2x^{2}+2x+3=0$$

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 2 for a, 2 for b, and 3 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\times 3}}{2\times 2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-20}}{4}$$

Solve the equation 2x^{2}+2x+3=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{5}i-1}{2} x=\frac{-1+\sqrt{5}i}{2}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{5}i-1}{2} x=\frac{-1+\sqrt{5}i}{2}$$

Solve for x

$$x=1$$

Solution Steps

Use the distributive property to multiply 2x^{2}+1 by x.

$$3=2x^{3}+x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{3}+x=3$$

Subtract 3 from both sides.

$$2x^{3}+x-3=0$$

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 2. List all candidates \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=1$$

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

$$2x^{2}+2x+3=0$$

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 2 for a, 2 for b, and 3 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\times 3}}{2\times 2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-20}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1$$