Solve (x^2-4x+3)*sqrt{2x^2-5x+2}=0

$$( x ^ { 2 } - 4 x + 3 ) \cdot \sqrt { 2 x ^ { 2 } - 5 x + 2 } = 0$$

Solve for x (complex solution)

$$x=1$$

$$x=3$$

$$x=2$$

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

Solution Steps

Use the distributive property to multiply x^{2}-4x+3 by \sqrt{2x^{2}-5x+2}.

$$x^{2}\sqrt{2x^{2}-5x+2}-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}=0$$

Subtract -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2} from both sides of the equation.

$$x^{2}\sqrt{2x^{2}-5x+2}=-\left(-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}\right)$$

To find the opposite of -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}, find the opposite of each term.

$$x^{2}\sqrt{2x^{2}-5x+2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}$$

Square both sides of the equation.

$$\left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Expand \left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}.

$$\left(x^{2}\right)^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

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

$$x^{4}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$x^{4}\left(2x^{2}-5x+2\right)=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use the distributive property to multiply x^{4} by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\sqrt{2x^{2}-5x+2}\sqrt{2x^{2}-5x+2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Multiply \sqrt{2x^{2}-5x+2} and \sqrt{2x^{2}-5x+2} to get \left(\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(2x^{2}-5x+2\right)-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use the distributive property to multiply 16x^{2} by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(2x^{2}-5x+2\right)+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use the distributive property to multiply -24x by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-48x^{3}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Combine -80x^{3} and -48x^{3} to get -128x^{3}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+32x^{2}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Combine 32x^{2} and 120x^{2} to get 152x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(2x^{2}-5x+2\right)$$

Use the distributive property to multiply 9 by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+18x^{2}-45x+18$$

Combine 152x^{2} and 18x^{2} to get 170x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-48x-45x+18$$

Combine -48x and -45x to get -93x.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-93x+18$$

Subtract 32x^{4} from both sides.

$$2x^{6}-5x^{5}+2x^{4}-32x^{4}=-128x^{3}+170x^{2}-93x+18$$

Combine 2x^{4} and -32x^{4} to get -30x^{4}.

$$2x^{6}-5x^{5}-30x^{4}=-128x^{3}+170x^{2}-93x+18$$

Add 128x^{3} to both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}=170x^{2}-93x+18$$

Subtract 170x^{2} from both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}=-93x+18$$

Add 93x to both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x=18$$

Subtract 18 from both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18 by x-1 to get 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18. Solve the equation where the result equals to 0.

$$2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 18 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18 by x-2 to get 2x^{4}+x^{3}-31x^{2}+33x-9. Solve the equation where the result equals to 0.

$$2x^{4}+x^{3}-31x^{2}+33x-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+x^{3}-31x^{2}+33x-9 by x-3 to get 2x^{3}+7x^{2}-10x+3. Solve the equation where the result equals to 0.

$$2x^{3}+7x^{2}-10x+3=0$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+7x^{2}-10x+3 by 2\left(x-\frac{1}{2}\right)=2x-1 to get x^{2}+4x-3. Solve the equation where the result equals to 0.

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

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 1 for a, 4 for b, and -3 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{-4±2\sqrt{7}}{2}$$

Solve the equation x^{2}+4x-3=0 when ± is plus and when ± is minus.

$$x=-\sqrt{7}-2 x=\sqrt{7}-2$$

List all found solutions.

$$x=1 x=2 x=3 x=\frac{1}{2} x=-\sqrt{7}-2 x=\sqrt{7}-2$$

Substitute 1 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(1^{2}-4+3\right)\sqrt{2\times 1^{2}-5+2}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute 2 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(2^{2}-4\times 2+3\right)\sqrt{2\times 2^{2}-5\times 2+2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute 3 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(3^{2}-4\times 3+3\right)\sqrt{2\times 3^{2}-5\times 3+2}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute \frac{1}{2} for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(\frac{1}{2}\right)^{2}-4\times \frac{1}{2}+3\right)\sqrt{2\times \left(\frac{1}{2}\right)^{2}-5\times \frac{1}{2}+2}=0$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$0=0$$

Substitute -\sqrt{7}-2 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(-\sqrt{7}-2\right)^{2}-4\left(-\sqrt{7}-2\right)+3\right)\sqrt{2\left(-\sqrt{7}-2\right)^{2}-5\left(-\sqrt{7}-2\right)+2}=0$$

Simplify. The value x=-\sqrt{7}-2 does not satisfy the equation.

$$\left(63720+24084\times 7^{\frac{1}{2}}\right)^{\frac{1}{2}}=0$$

Substitute \sqrt{7}-2 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(\sqrt{7}-2\right)^{2}-4\left(\sqrt{7}-2\right)+3\right)\sqrt{2\left(\sqrt{7}-2\right)^{2}-5\left(\sqrt{7}-2\right)+2}=0$$

Simplify. The value x=\sqrt{7}-2 does not satisfy the equation.

$$i\left(-\left(63720-24084\times 7^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}=0$$

List all solutions of \sqrt{2x^{2}-5x+2}x^{2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}.

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

Solve for x

$$x=3$$

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

$$x=2$$

Solution Steps

Use the distributive property to multiply x^{2}-4x+3 by \sqrt{2x^{2}-5x+2}.

$$x^{2}\sqrt{2x^{2}-5x+2}-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}=0$$

Subtract -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2} from both sides of the equation.

$$x^{2}\sqrt{2x^{2}-5x+2}=-\left(-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}\right)$$

To find the opposite of -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}, find the opposite of each term.

$$x^{2}\sqrt{2x^{2}-5x+2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}$$

Square both sides of the equation.

$$\left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Expand \left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}.

$$\left(x^{2}\right)^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

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

$$x^{4}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$x^{4}\left(2x^{2}-5x+2\right)=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use the distributive property to multiply x^{4} by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\sqrt{2x^{2}-5x+2}\sqrt{2x^{2}-5x+2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Multiply \sqrt{2x^{2}-5x+2} and \sqrt{2x^{2}-5x+2} to get \left(\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(2x^{2}-5x+2\right)-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use the distributive property to multiply 16x^{2} by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(2x^{2}-5x+2\right)+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Use the distributive property to multiply -24x by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-48x^{3}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Combine -80x^{3} and -48x^{3} to get -128x^{3}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+32x^{2}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Combine 32x^{2} and 120x^{2} to get 152x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Calculate \sqrt{2x^{2}-5x+2} to the power of 2 and get 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(2x^{2}-5x+2\right)$$

Use the distributive property to multiply 9 by 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+18x^{2}-45x+18$$

Combine 152x^{2} and 18x^{2} to get 170x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-48x-45x+18$$

Combine -48x and -45x to get -93x.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-93x+18$$

Subtract 32x^{4} from both sides.

$$2x^{6}-5x^{5}+2x^{4}-32x^{4}=-128x^{3}+170x^{2}-93x+18$$

Combine 2x^{4} and -32x^{4} to get -30x^{4}.

$$2x^{6}-5x^{5}-30x^{4}=-128x^{3}+170x^{2}-93x+18$$

Add 128x^{3} to both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}=170x^{2}-93x+18$$

Subtract 170x^{2} from both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}=-93x+18$$

Add 93x to both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x=18$$

Subtract 18 from both sides.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18 by x-1 to get 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18. Solve the equation where the result equals to 0.

$$2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 18 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18 by x-2 to get 2x^{4}+x^{3}-31x^{2}+33x-9. Solve the equation where the result equals to 0.

$$2x^{4}+x^{3}-31x^{2}+33x-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+x^{3}-31x^{2}+33x-9 by x-3 to get 2x^{3}+7x^{2}-10x+3. Solve the equation where the result equals to 0.

$$2x^{3}+7x^{2}-10x+3=0$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+7x^{2}-10x+3 by 2\left(x-\frac{1}{2}\right)=2x-1 to get x^{2}+4x-3. Solve the equation where the result equals to 0.

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

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 1 for a, 4 for b, and -3 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{-4±2\sqrt{7}}{2}$$

Solve the equation x^{2}+4x-3=0 when ± is plus and when ± is minus.

$$x=-\sqrt{7}-2 x=\sqrt{7}-2$$

List all found solutions.

$$x=1 x=2 x=3 x=\frac{1}{2} x=-\sqrt{7}-2 x=\sqrt{7}-2$$

Substitute 1 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0. The expression \sqrt{2\times 1^{2}-5+2} is undefined because the radicand cannot be negative.

$$\left(1^{2}-4+3\right)\sqrt{2\times 1^{2}-5+2}=0$$

Substitute 2 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(2^{2}-4\times 2+3\right)\sqrt{2\times 2^{2}-5\times 2+2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute 3 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(3^{2}-4\times 3+3\right)\sqrt{2\times 3^{2}-5\times 3+2}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute \frac{1}{2} for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(\frac{1}{2}\right)^{2}-4\times \frac{1}{2}+3\right)\sqrt{2\times \left(\frac{1}{2}\right)^{2}-5\times \frac{1}{2}+2}=0$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$0=0$$

Substitute -\sqrt{7}-2 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(-\sqrt{7}-2\right)^{2}-4\left(-\sqrt{7}-2\right)+3\right)\sqrt{2\left(-\sqrt{7}-2\right)^{2}-5\left(-\sqrt{7}-2\right)+2}=0$$

Simplify. The value x=-\sqrt{7}-2 does not satisfy the equation.

$$\left(63720+24084\times 7^{\frac{1}{2}}\right)^{\frac{1}{2}}=0$$

Substitute \sqrt{7}-2 for x in the equation \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0. The expression \sqrt{2\left(\sqrt{7}-2\right)^{2}-5\left(\sqrt{7}-2\right)+2} is undefined because the radicand cannot be negative.

$$\left(\left(\sqrt{7}-2\right)^{2}-4\left(\sqrt{7}-2\right)+3\right)\sqrt{2\left(\sqrt{7}-2\right)^{2}-5\left(\sqrt{7}-2\right)+2}=0$$

List all solutions of \sqrt{2x^{2}-5x+2}x^{2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}.

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