Solve x^4=(3x-10)^2

$$x ^ { 4 } = ( 3 x - 10 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-5$$

$$x=2$$

$$x=\frac{-\sqrt{31}i+3}{2}\approx 1.5-2.783882181i$$

$$x=\frac{3+\sqrt{31}i}{2}\approx 1.5+2.783882181i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-10\right)^{2}.

$$x^{4}=9x^{2}-60x+100$$

Subtract 9x^{2} from both sides.

$$x^{4}-9x^{2}=-60x+100$$

Add 60x to both sides.

$$x^{4}-9x^{2}+60x=100$$

Subtract 100 from both sides.

$$x^{4}-9x^{2}+60x-100=0$$

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 1. List all candidates \frac{p}{q}.

$$±100,±50,±25,±20,±10,±5,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-9x^{2}+60x-100 by x-2 to get x^{3}+2x^{2}-5x+50. Solve the equation where the result equals to 0.

$$x^{3}+2x^{2}-5x+50=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 50 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±50,±25,±10,±5,±2,±1$$

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.

$$x=-5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}-5x+50 by x+5 to get x^{2}-3x+10. Solve the equation where the result equals to 0.

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

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 10 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-31}}{2}$$

Solve the equation x^{2}-3x+10=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{31}i+3}{2} x=\frac{3+\sqrt{31}i}{2}$$

List all found solutions.

$$x=2 x=-5 x=\frac{-\sqrt{31}i+3}{2} x=\frac{3+\sqrt{31}i}{2}$$

Solve for x

$$x=-5$$

$$x=2$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-10\right)^{2}.

$$x^{4}=9x^{2}-60x+100$$

Subtract 9x^{2} from both sides.

$$x^{4}-9x^{2}=-60x+100$$

Add 60x to both sides.

$$x^{4}-9x^{2}+60x=100$$

Subtract 100 from both sides.

$$x^{4}-9x^{2}+60x-100=0$$

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 1. List all candidates \frac{p}{q}.

$$±100,±50,±25,±20,±10,±5,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-9x^{2}+60x-100 by x-2 to get x^{3}+2x^{2}-5x+50. Solve the equation where the result equals to 0.

$$x^{3}+2x^{2}-5x+50=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 50 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±50,±25,±10,±5,±2,±1$$

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.

$$x=-5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}-5x+50 by x+5 to get x^{2}-3x+10. Solve the equation where the result equals to 0.

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

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 10 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-31}}{2}$$

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.

$$x=2 x=-5$$