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

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

Solve for x

$$x=1$$

$$x = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

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

$$1=x^{2}-2x+1+4x^{2}-12x+9$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$1=5x^{2}-2x+1-12x+9$$

Combine -2x and -12x to get -14x.

$$1=5x^{2}-14x+1+9$$

Add 1 and 9 to get 10.

$$1=5x^{2}-14x+10$$

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

$$5x^{2}-14x+10=1$$

Subtract 1 from both sides.

$$5x^{2}-14x+10-1=0$$

Subtract 1 from 10 to get 9.

$$5x^{2}-14x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-14 ab=5\times 9=45$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite 5x^{2}-14x+9 as \left(5x^{2}-9x\right)+\left(-5x+9\right).

$$\left(5x^{2}-9x\right)+\left(-5x+9\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(5x-9\right)-\left(5x-9\right)$$

Factor out common term 5x-9 by using distributive property.

$$\left(5x-9\right)\left(x-1\right)$$

To find equation solutions, solve 5x-9=0 and x-1=0.

$$x=\frac{9}{5} x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

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

$$1=x^{2}-2x+1+4x^{2}-12x+9$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$1=5x^{2}-2x+1-12x+9$$

Combine -2x and -12x to get -14x.

$$1=5x^{2}-14x+1+9$$

Add 1 and 9 to get 10.

$$1=5x^{2}-14x+10$$

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

$$5x^{2}-14x+10=1$$

Subtract 1 from both sides.

$$5x^{2}-14x+10-1=0$$

Subtract 1 from 10 to get 9.

$$5x^{2}-14x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -14 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 5\times 9}}{2\times 5}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 5\times 9}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-14\right)±\sqrt{196-20\times 9}}{2\times 5}$$

Multiply -20 times 9.

$$x=\frac{-\left(-14\right)±\sqrt{196-180}}{2\times 5}$$

Add 196 to -180.

$$x=\frac{-\left(-14\right)±\sqrt{16}}{2\times 5}$$

Take the square root of 16.

$$x=\frac{-\left(-14\right)±4}{2\times 5}$$

The opposite of -14 is 14.

$$x=\frac{14±4}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{14±4}{10}$$

Now solve the equation x=\frac{14±4}{10} when ± is plus. Add 14 to 4.

$$x=\frac{18}{10}$$

Reduce the fraction \frac{18}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{5}$$

Now solve the equation x=\frac{14±4}{10} when ± is minus. Subtract 4 from 14.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

The equation is now solved.

$$x=\frac{9}{5} x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

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

$$1=x^{2}-2x+1+4x^{2}-12x+9$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$1=5x^{2}-2x+1-12x+9$$

Combine -2x and -12x to get -14x.

$$1=5x^{2}-14x+1+9$$

Add 1 and 9 to get 10.

$$1=5x^{2}-14x+10$$

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

$$5x^{2}-14x+10=1$$

Subtract 10 from both sides.

$$5x^{2}-14x=1-10$$

Subtract 10 from 1 to get -9.

$$5x^{2}-14x=-9$$

Divide both sides by 5.

$$\frac{5x^{2}-14x}{5}=-\frac{9}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{14}{5}x=-\frac{9}{5}$$

Divide -\frac{14}{5}, the coefficient of the x term, by 2 to get -\frac{7}{5}. Then add the square of -\frac{7}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{5}x+\left(-\frac{7}{5}\right)^{2}=-\frac{9}{5}+\left(-\frac{7}{5}\right)^{2}$$

Square -\frac{7}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{5}x+\frac{49}{25}=-\frac{9}{5}+\frac{49}{25}$$

Add -\frac{9}{5} to \frac{49}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{14}{5}x+\frac{49}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{14}{5}x+\frac{49}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$x-\frac{7}{5}=\frac{2}{5} x-\frac{7}{5}=-\frac{2}{5}$$

Add \frac{7}{5} to both sides of the equation.

$$x=\frac{9}{5} x=1$$