Solve (2x^2-3x)*(2x^2-3x+5)=14

$$( 2 x ^ { 2 } - 3 x ) \cdot ( 2 x ^ { 2 } - 3 x + 5 ) = 14$$

Solve for x (complex solution)

$$x=2$$

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

$$x=\frac{3+\sqrt{47}i}{4}\approx 0.75+1.71391365i$$

$$x=\frac{-\sqrt{47}i+3}{4}\approx 0.75-1.71391365i$$

Solution Steps

Use the distributive property to multiply 2x^{2}-3x by 2x^{2}-3x+5 and combine like terms.

$$4x^{4}-12x^{3}+19x^{2}-15x=14$$

Subtract 14 from both sides.

$$4x^{4}-12x^{3}+19x^{2}-15x-14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -14 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{7}{2},±7,±14,±\frac{7}{4},±\frac{1}{2},±1,±2,±\frac{1}{4}$$

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 4x^{4}-12x^{3}+19x^{2}-15x-14 by x-2 to get 4x^{3}-4x^{2}+11x+7. Solve the equation where the result equals to 0.

$$4x^{3}-4x^{2}+11x+7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 7 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{7}{4},±\frac{7}{2},±7,±\frac{1}{4},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-4x^{2}+11x+7 by 2\left(x+\frac{1}{2}\right)=2x+1 to get 2x^{2}-3x+7. Solve the equation where the result equals to 0.

$$2x^{2}-3x+7=0$$

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, -3 for b, and 7 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 7}}{2\times 2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-47}}{4}$$

Solve the equation 2x^{2}-3x+7=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{47}i+3}{4} x=\frac{3+\sqrt{47}i}{4}$$

List all found solutions.

$$x=2 x=-\frac{1}{2} x=\frac{-\sqrt{47}i+3}{4} x=\frac{3+\sqrt{47}i}{4}$$

Solve for x

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

$$x=2$$

Solution Steps

Use the distributive property to multiply 2x^{2}-3x by 2x^{2}-3x+5 and combine like terms.

$$4x^{4}-12x^{3}+19x^{2}-15x=14$$

Subtract 14 from both sides.

$$4x^{4}-12x^{3}+19x^{2}-15x-14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -14 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{7}{2},±7,±14,±\frac{7}{4},±\frac{1}{2},±1,±2,±\frac{1}{4}$$

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 4x^{4}-12x^{3}+19x^{2}-15x-14 by x-2 to get 4x^{3}-4x^{2}+11x+7. Solve the equation where the result equals to 0.

$$4x^{3}-4x^{2}+11x+7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 7 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{7}{4},±\frac{7}{2},±7,±\frac{1}{4},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-4x^{2}+11x+7 by 2\left(x+\frac{1}{2}\right)=2x+1 to get 2x^{2}-3x+7. Solve the equation where the result equals to 0.

$$2x^{2}-3x+7=0$$

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, -3 for b, and 7 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 7}}{2\times 2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-47}}{4}$$

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.

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