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

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

Solve for x

$$x=1$$

$$x=2$$

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

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

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x^{2}-3x\right)^{2}.

$$4\left(x^{2}\right)^{2}-12x^{2}x+9x^{2}-2x^{2}+3x=2$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$4x^{4}-12x^{2}x+9x^{2}-2x^{2}+3x=2$$

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

$$4x^{4}-12x^{3}+9x^{2}-2x^{2}+3x=2$$

Combine 9x^{2} and -2x^{2} to get 7x^{2}.

$$4x^{4}-12x^{3}+7x^{2}+3x=2$$

Subtract 2 from both sides.

$$4x^{4}-12x^{3}+7x^{2}+3x-2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1,±2,±\frac{1}{4}$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{4}-12x^{3}+7x^{2}+3x-2 by x-1 to get 4x^{3}-8x^{2}-x+2. Solve the equation where the result equals to 0.

$$4x^{3}-8x^{2}-x+2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1,±2,±\frac{1}{4}$$

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^{3}-8x^{2}-x+2 by x-2 to get 4x^{2}-1. Solve the equation where the result equals to 0.

$$4x^{2}-1=0$$

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 4 for a, 0 for b, and -1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{0±4}{8}$$

Solve the equation 4x^{2}-1=0 when ± is plus and when ± is minus.

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

List all found solutions.

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