Solve (x^2+3x)^2+(x^2+3x)-20=0

$$( x ^ { 2 } + 3 x ) ^ { 2 } + ( x ^ { 2 } + 3 x ) - 20 = 0$$

Solve for x (complex solution)

$$x=1$$

$$x=-4$$

$$x=\frac{-3+\sqrt{11}i}{2}\approx -1.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i-3}{2}\approx -1.5-1.658312395i$$

Solution Steps

Simplify.

$$x^{4}+6x^{3}+10x^{2}+3x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+10x^{2}+3x-20 by x-1 to get x^{3}+7x^{2}+17x+20. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}+17x+20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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 x^{3}+7x^{2}+17x+20 by x+4 to get x^{2}+3x+5. Solve the equation where the result equals to 0.

$$x^{2}+3x+5=0$$

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 1 for a, 3 for b, and 5 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-11}}{2}$$

Solve the equation x^{2}+3x+5=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{11}i-3}{2} x=\frac{-3+\sqrt{11}i}{2}$$

List all found solutions.

$$x=1 x=-4 x=\frac{-\sqrt{11}i-3}{2} x=\frac{-3+\sqrt{11}i}{2}$$

Solve for x

$$x=-4$$

$$x=1$$

Solution Steps

Simplify.

$$x^{4}+6x^{3}+10x^{2}+3x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+10x^{2}+3x-20 by x-1 to get x^{3}+7x^{2}+17x+20. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}+17x+20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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 x^{3}+7x^{2}+17x+20 by x+4 to get x^{2}+3x+5. Solve the equation where the result equals to 0.

$$x^{2}+3x+5=0$$

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 1 for a, 3 for b, and 5 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-11}}{2}$$

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.

$$x=1 x=-4$$