Solve 1^2+(a+3/2)^2+1+(2-a)^2=49/4

$$1 ^ { 2 } + ( a + \frac { 3 } { 2 } ) ^ { 2 } + 1 + ( 2 - a ) ^ { 2 } = \frac { 49 } { 4 }$$

Solve for a

$$a = \frac{\sqrt{33} + 1}{4} \approx 1.686140662$$

$$a=\frac{1-\sqrt{33}}{4}\approx -1.186140662$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 2 and get 1.

$$1+\left(a+\frac{3}{2}\right)^{2}+1+\left(2-a\right)^{2}=\frac{49}{4}$$

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

$$1+a^{2}+3a+\frac{9}{4}+1+\left(2-a\right)^{2}=\frac{49}{4}$$

Add 1 and \frac{9}{4} to get \frac{13}{4}.

$$\frac{13}{4}+a^{2}+3a+1+\left(2-a\right)^{2}=\frac{49}{4}$$

Add \frac{13}{4} and 1 to get \frac{17}{4}.

$$\frac{17}{4}+a^{2}+3a+\left(2-a\right)^{2}=\frac{49}{4}$$

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

$$\frac{17}{4}+a^{2}+3a+4-4a+a^{2}=\frac{49}{4}$$

Add \frac{17}{4} and 4 to get \frac{33}{4}.

$$\frac{33}{4}+a^{2}+3a-4a+a^{2}=\frac{49}{4}$$

Combine 3a and -4a to get -a.

$$\frac{33}{4}+a^{2}-a+a^{2}=\frac{49}{4}$$

Combine a^{2} and a^{2} to get 2a^{2}.

$$\frac{33}{4}+2a^{2}-a=\frac{49}{4}$$

Subtract \frac{49}{4} from both sides.

$$\frac{33}{4}+2a^{2}-a-\frac{49}{4}=0$$

Subtract \frac{49}{4} from \frac{33}{4} to get -4.

$$-4+2a^{2}-a=0$$

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.

$$2a^{2}-a-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Multiply -4 times 2.

$$a=\frac{-\left(-1\right)±\sqrt{1-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$a=\frac{-\left(-1\right)±\sqrt{1+32}}{2\times 2}$$

Add 1 to 32.

$$a=\frac{-\left(-1\right)±\sqrt{33}}{2\times 2}$$

The opposite of -1 is 1.

$$a=\frac{1±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{1±\sqrt{33}}{4}$$

Now solve the equation a=\frac{1±\sqrt{33}}{4} when ± is plus. Add 1 to \sqrt{33}.

$$a=\frac{\sqrt{33}+1}{4}$$

Now solve the equation a=\frac{1±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from 1.

$$a=\frac{1-\sqrt{33}}{4}$$

The equation is now solved.

$$a=\frac{\sqrt{33}+1}{4} a=\frac{1-\sqrt{33}}{4}$$

Steps for Completing the Square

Calculate 1 to the power of 2 and get 1.

$$1+\left(a+\frac{3}{2}\right)^{2}+1+\left(2-a\right)^{2}=\frac{49}{4}$$

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

$$1+a^{2}+3a+\frac{9}{4}+1+\left(2-a\right)^{2}=\frac{49}{4}$$

Add 1 and \frac{9}{4} to get \frac{13}{4}.

$$\frac{13}{4}+a^{2}+3a+1+\left(2-a\right)^{2}=\frac{49}{4}$$

Add \frac{13}{4} and 1 to get \frac{17}{4}.

$$\frac{17}{4}+a^{2}+3a+\left(2-a\right)^{2}=\frac{49}{4}$$

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

$$\frac{17}{4}+a^{2}+3a+4-4a+a^{2}=\frac{49}{4}$$

Add \frac{17}{4} and 4 to get \frac{33}{4}.

$$\frac{33}{4}+a^{2}+3a-4a+a^{2}=\frac{49}{4}$$

Combine 3a and -4a to get -a.

$$\frac{33}{4}+a^{2}-a+a^{2}=\frac{49}{4}$$

Combine a^{2} and a^{2} to get 2a^{2}.

$$\frac{33}{4}+2a^{2}-a=\frac{49}{4}$$

Subtract \frac{33}{4} from both sides.

$$2a^{2}-a=\frac{49}{4}-\frac{33}{4}$$

Subtract \frac{33}{4} from \frac{49}{4} to get 4.

$$2a^{2}-a=4$$

Divide both sides by 2.

$$\frac{2a^{2}-a}{2}=\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}-\frac{1}{2}a=\frac{4}{2}$$

Divide 4 by 2.

$$a^{2}-\frac{1}{2}a=2$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{1}{2}a+\left(-\frac{1}{4}\right)^{2}=2+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=2+\frac{1}{16}$$

Add 2 to \frac{1}{16}.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{33}{16}$$

Factor a^{2}-\frac{1}{2}a+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{1}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$a-\frac{1}{4}=\frac{\sqrt{33}}{4} a-\frac{1}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$a=\frac{\sqrt{33}+1}{4} a=\frac{1-\sqrt{33}}{4}$$