Solve {left((x-1)*sqrt{2}right)}^2+{left(sqrt{2-3x}right)}^2

$${ \left((x-1) \times \sqrt{ 2 } \right) }^{ 2 } + { \left( \sqrt{ 2-3x } \right) }^{ 2 }$$

Evaluate

$$2x^{2}-7x+4$$

Solution Steps

Use the distributive property to multiply x-1 by \sqrt{2}.

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

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

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$x^{2}\left(\sqrt{2}\right)^{2}-2x\times 2+\left(\sqrt{2}\right)^{2}+\left(\sqrt{2-3x}\right)^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}\times 2-2x\times 2+\left(\sqrt{2}\right)^{2}+\left(\sqrt{2-3x}\right)^{2}$$

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Calculate \sqrt{2-3x} to the power of 2 and get 2-3x.

$$x^{2}\times 2-4x+2+2-3x$$

Add 2 and 2 to get 4.

$$x^{2}\times 2-4x+4-3x$$

Combine -4x and -3x to get -7x.

$$x^{2}\times 2-7x+4$$

Factor

$$2\left(x-\frac{7-\sqrt{17}}{4}\right)\left(x-\frac{\sqrt{17}+7}{4}\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-1 by \sqrt{2}.

$$factor(\left(x\sqrt{2}-\sqrt{2}\right)^{2}+\left(\sqrt{2-3x}\right)^{2})$$

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

$$factor(x^{2}\left(\sqrt{2}\right)^{2}-2x\sqrt{2}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{2-3x}\right)^{2})$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$factor(x^{2}\left(\sqrt{2}\right)^{2}-2x\times 2+\left(\sqrt{2}\right)^{2}+\left(\sqrt{2-3x}\right)^{2})$$

The square of \sqrt{2} is 2.

$$factor(x^{2}\times 2-2x\times 2+\left(\sqrt{2}\right)^{2}+\left(\sqrt{2-3x}\right)^{2})$$

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Calculate \sqrt{2-3x} to the power of 2 and get 2-3x.

$$factor(x^{2}\times 2-4x+2+2-3x)$$

Add 2 and 2 to get 4.

$$factor(x^{2}\times 2-4x+4-3x)$$

Combine -4x and -3x to get -7x.

$$factor(x^{2}\times 2-7x+4)$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 4}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-\left(-7\right)±\sqrt{49-32}}{2\times 2}$$

Add 49 to -32.

$$x=\frac{-\left(-7\right)±\sqrt{17}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{17}}{4}$$

Now solve the equation x=\frac{7±\sqrt{17}}{4} when ± is plus. Add 7 to \sqrt{17}.

$$x=\frac{\sqrt{17}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from 7.

$$x=\frac{7-\sqrt{17}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7+\sqrt{17}}{4} for x_{1} and \frac{7-\sqrt{17}}{4} for x_{2}.

$$2x^{2}-7x+4=2\left(x-\frac{\sqrt{17}+7}{4}\right)\left(x-\frac{7-\sqrt{17}}{4}\right)$$