Solve (x+1)^2+(x^2/4-1/2)^2=5/4

$$( x + 1 ) ^ { 2 } + ( \frac { x ^ { 2 } } { 4 } - \frac { 1 } { 2 } ) ^ { 2 } = \frac { 5 } { 4 }$$

Solve for x (complex solution)

$$x=-2$$

$$x=0$$

$$x=-\sqrt{15}i+1\approx 1-3.872983346i$$

$$x=1+\sqrt{15}i\approx 1+3.872983346i$$

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}.

$$x^{2}+2x+1+\left(\frac{x^{2}}{4}-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

$$x^{2}+2x+1+\left(\frac{x^{2}}{4}-\frac{2}{4}\right)^{2}=\frac{5}{4}$$

Since \frac{x^{2}}{4} and \frac{2}{4} have the same denominator, subtract them by subtracting their numerators.

$$x^{2}+2x+1+\left(\frac{x^{2}-2}{4}\right)^{2}=\frac{5}{4}$$

To raise \frac{x^{2}-2}{4} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}+2x+1+\frac{\left(x^{2}-2\right)^{2}}{4^{2}}=\frac{5}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+2x+1 times \frac{4^{2}}{4^{2}}.

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

Since \frac{\left(x^{2}+2x+1\right)\times 4^{2}}{4^{2}} and \frac{\left(x^{2}-2\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(x^{2}+2x+1\right)\times 4^{2}+\left(x^{2}-2\right)^{2}.

$$\frac{16x^{2}+32x+16+x^{4}-4x^{2}+4}{4^{2}}=\frac{5}{4}$$

Combine like terms in 16x^{2}+32x+16+x^{4}-4x^{2}+4.

$$\frac{12x^{2}+32x+20+x^{4}}{4^{2}}=\frac{5}{4}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{12x^{2}+32x+20+x^{4}}{16}=\frac{5}{4}$$

Divide each term of 12x^{2}+32x+20+x^{4} by 16 to get \frac{3}{4}x^{2}+2x+\frac{5}{4}+\frac{1}{16}x^{4}.

$$\frac{3}{4}x^{2}+2x+\frac{5}{4}+\frac{1}{16}x^{4}=\frac{5}{4}$$

Subtract \frac{5}{4} from both sides.

$$\frac{3}{4}x^{2}+2x+\frac{5}{4}+\frac{1}{16}x^{4}-\frac{5}{4}=0$$

Subtract \frac{5}{4} from \frac{5}{4} to get 0.

$$\frac{3}{4}x^{2}+2x+\frac{1}{16}x^{4}=0$$

Substitute t for x^{2}.

$$\frac{1}{16}t^{2}+\frac{3}{4}t+2=0$$

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 \frac{1}{16} for a, \frac{3}{4} for b, and 2 for c in the quadratic formula.

$$t=\frac{-\frac{3}{4}±\sqrt{\left(\frac{3}{4}\right)^{2}-4\times \frac{1}{16}\times 2}}{\frac{1}{16}\times 2}$$

Do the calculations.

$$t=\frac{-\frac{3}{4}±\frac{1}{4}}{\frac{1}{8}}$$

Solve the equation t=\frac{-\frac{3}{4}±\frac{1}{4}}{\frac{1}{8}} when ± is plus and when ± is minus.

$$t=-4 t=-8$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

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

Solve for x

$$x=-2$$

$$x=0$$

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}.

$$x^{2}+2x+1+\left(\frac{x^{2}}{4}-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

$$x^{2}+2x+1+\left(\frac{x^{2}}{4}-\frac{2}{4}\right)^{2}=\frac{5}{4}$$

Since \frac{x^{2}}{4} and \frac{2}{4} have the same denominator, subtract them by subtracting their numerators.

$$x^{2}+2x+1+\left(\frac{x^{2}-2}{4}\right)^{2}=\frac{5}{4}$$

To raise \frac{x^{2}-2}{4} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}+2x+1+\frac{\left(x^{2}-2\right)^{2}}{4^{2}}=\frac{5}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+2x+1 times \frac{4^{2}}{4^{2}}.

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

Since \frac{\left(x^{2}+2x+1\right)\times 4^{2}}{4^{2}} and \frac{\left(x^{2}-2\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(x^{2}+2x+1\right)\times 4^{2}+\left(x^{2}-2\right)^{2}.

$$\frac{16x^{2}+32x+16+x^{4}-4x^{2}+4}{4^{2}}=\frac{5}{4}$$

Combine like terms in 16x^{2}+32x+16+x^{4}-4x^{2}+4.

$$\frac{12x^{2}+32x+20+x^{4}}{4^{2}}=\frac{5}{4}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{12x^{2}+32x+20+x^{4}}{16}=\frac{5}{4}$$

Divide each term of 12x^{2}+32x+20+x^{4} by 16 to get \frac{3}{4}x^{2}+2x+\frac{5}{4}+\frac{1}{16}x^{4}.

$$\frac{3}{4}x^{2}+2x+\frac{5}{4}+\frac{1}{16}x^{4}=\frac{5}{4}$$

Subtract \frac{5}{4} from both sides.

$$\frac{3}{4}x^{2}+2x+\frac{5}{4}+\frac{1}{16}x^{4}-\frac{5}{4}=0$$

Subtract \frac{5}{4} from \frac{5}{4} to get 0.

$$\frac{3}{4}x^{2}+2x+\frac{1}{16}x^{4}=0$$

Substitute t for x^{2}.

$$\frac{1}{16}t^{2}+\frac{3}{4}t+2=0$$

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 \frac{1}{16} for a, \frac{3}{4} for b, and 2 for c in the quadratic formula.

$$t=\frac{-\frac{3}{4}±\sqrt{\left(\frac{3}{4}\right)^{2}-4\times \frac{1}{16}\times 2}}{\frac{1}{16}\times 2}$$

Do the calculations.

$$t=\frac{-\frac{3}{4}±\frac{1}{4}}{\frac{1}{8}}$$

Solve the equation t=\frac{-\frac{3}{4}±\frac{1}{4}}{\frac{1}{8}} when ± is plus and when ± is minus.

$$t=-4 t=-8$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x\in \emptyset$$