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

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{79}i}{4}\approx 0.25+2.222048604i$$

$$x=\frac{-\sqrt{79}i+1}{4}\approx 0.25-2.222048604i$$

$$x=2$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Solution Steps

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

$$4x^{4}=100-20x+x^{2}$$

Subtract 100 from both sides.

$$4x^{4}-100=-20x+x^{2}$$

Add 20x to both sides.

$$4x^{4}-100+20x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{4}-100+20x-x^{2}=0$$

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

$$4x^{4}-x^{2}+20x-100=0$$

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

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

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 4x^{4}-x^{2}+20x-100 by x-2 to get 4x^{3}+8x^{2}+15x+50. Solve the equation where the result equals to 0.

$$4x^{3}+8x^{2}+15x+50=0$$

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

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

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.

$$x=-\frac{5}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+8x^{2}+15x+50 by 2\left(x+\frac{5}{2}\right)=2x+5 to get 2x^{2}-x+10. Solve the equation where the result equals to 0.

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

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 2 for a, -1 for b, and 10 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{1±\sqrt{-79}}{4}$$

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

$$x=\frac{-\sqrt{79}i+1}{4} x=\frac{1+\sqrt{79}i}{4}$$

List all found solutions.

$$x=2 x=-\frac{5}{2} x=\frac{-\sqrt{79}i+1}{4} x=\frac{1+\sqrt{79}i}{4}$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=2$$

Solution Steps

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

$$4x^{4}=100-20x+x^{2}$$

Subtract 100 from both sides.

$$4x^{4}-100=-20x+x^{2}$$

Add 20x to both sides.

$$4x^{4}-100+20x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{4}-100+20x-x^{2}=0$$

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

$$4x^{4}-x^{2}+20x-100=0$$

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

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

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 4x^{4}-x^{2}+20x-100 by x-2 to get 4x^{3}+8x^{2}+15x+50. Solve the equation where the result equals to 0.

$$4x^{3}+8x^{2}+15x+50=0$$

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

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

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.

$$x=-\frac{5}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+8x^{2}+15x+50 by 2\left(x+\frac{5}{2}\right)=2x+5 to get 2x^{2}-x+10. Solve the equation where the result equals to 0.

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

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 2 for a, -1 for b, and 10 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{1±\sqrt{-79}}{4}$$

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.

$$x=2 x=-\frac{5}{2}$$