Solve (x^2-16)^2+(x^2+x-12)^2=0

$$( x ^ { 2 } - 16 ) ^ { 2 } + ( x ^ { 2 } + x - 12 ) ^ { 2 } = 0$$

Solve for x (complex solution)

$$x=\frac{7}{2}+\frac{1}{2}i=3.5+0.5i$$

$$x=\frac{7}{2}-\frac{1}{2}i=3.5-0.5i$$

$$x=-4$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$\left(x^{2}+x-12\right)^{2}+\left(x^{2}-16\right)^{2}=0$$

Simplify.

$$2x^{4}+2x^{3}-55x^{2}-24x+400=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 400 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±200,±400,±100,±50,±40,±80,±25,±20,±\frac{25}{2},±10,±8,±16,±5,±4,±\frac{5}{2},±2,±1,±\frac{1}{2}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+2x^{3}-55x^{2}-24x+400 by x+4 to get 2x^{3}-6x^{2}-31x+100. Solve the equation where the result equals to 0.

$$2x^{3}-6x^{2}-31x+100=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 100 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±50,±100,±25,±\frac{25}{2},±10,±20,±5,±\frac{5}{2},±2,±4,±1,±\frac{1}{2}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-6x^{2}-31x+100 by x+4 to get 2x^{2}-14x+25. Solve the equation where the result equals to 0.

$$2x^{2}-14x+25=0$$

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 2 for a, -14 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 25}}{2\times 2}$$

Do the calculations.

$$x=\frac{14±\sqrt{-4}}{4}$$

Solve the equation 2x^{2}-14x+25=0 when ± is plus and when ± is minus.

$$x=\frac{7}{2}-\frac{1}{2}i x=\frac{7}{2}+\frac{1}{2}i$$

List all found solutions.

$$x=-4 x=\frac{7}{2}-\frac{1}{2}i x=\frac{7}{2}+\frac{1}{2}i$$

Solve for x

$$x=-4$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$\left(x^{2}+x-12\right)^{2}+\left(x^{2}-16\right)^{2}=0$$

Simplify.

$$2x^{4}+2x^{3}-55x^{2}-24x+400=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 400 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±200,±400,±100,±50,±40,±80,±25,±20,±\frac{25}{2},±10,±8,±16,±5,±4,±\frac{5}{2},±2,±1,±\frac{1}{2}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+2x^{3}-55x^{2}-24x+400 by x+4 to get 2x^{3}-6x^{2}-31x+100. Solve the equation where the result equals to 0.

$$2x^{3}-6x^{2}-31x+100=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 100 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±50,±100,±25,±\frac{25}{2},±10,±20,±5,±\frac{5}{2},±2,±4,±1,±\frac{1}{2}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-6x^{2}-31x+100 by x+4 to get 2x^{2}-14x+25. Solve the equation where the result equals to 0.

$$2x^{2}-14x+25=0$$

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 2 for a, -14 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 25}}{2\times 2}$$

Do the calculations.

$$x=\frac{14±\sqrt{-4}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-4$$