Solve x^4=left(5x-14right)^2

$${ x }^{ 4 } = { \left(5x-14 \right) }^{ 2 }$$

Solve for x (complex solution)

$$x=-7$$

$$x=2$$

$$x=\frac{-\sqrt{31}i+5}{2}\approx 2.5-2.783882181i$$

$$x=\frac{5+\sqrt{31}i}{2}\approx 2.5+2.783882181i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-14\right)^{2}.

$$x^{4}=25x^{2}-140x+196$$

Subtract 25x^{2} from both sides.

$$x^{4}-25x^{2}=-140x+196$$

Add 140x to both sides.

$$x^{4}-25x^{2}+140x=196$$

Subtract 196 from both sides.

$$x^{4}-25x^{2}+140x-196=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -196 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±196,±98,±49,±28,±14,±7,±4,±2,±1$$

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}-25x^{2}+140x-196 by x-2 to get x^{3}+2x^{2}-21x+98. Solve the equation where the result equals to 0.

$$x^{3}+2x^{2}-21x+98=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 98 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±98,±49,±14,±7,±2,±1$$

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.

$$x=-7$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}-21x+98 by x+7 to get x^{2}-5x+14. Solve the equation where the result equals to 0.

$$x^{2}-5x+14=0$$

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, -5 for b, and 14 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 14}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{-31}}{2}$$

Solve the equation x^{2}-5x+14=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=2 x=-7 x=\frac{-\sqrt{31}i+5}{2} x=\frac{5+\sqrt{31}i}{2}$$

Solve for x

$$x=-7$$

$$x=2$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-14\right)^{2}.

$$x^{4}=25x^{2}-140x+196$$

Subtract 25x^{2} from both sides.

$$x^{4}-25x^{2}=-140x+196$$

Add 140x to both sides.

$$x^{4}-25x^{2}+140x=196$$

Subtract 196 from both sides.

$$x^{4}-25x^{2}+140x-196=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -196 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±196,±98,±49,±28,±14,±7,±4,±2,±1$$

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}-25x^{2}+140x-196 by x-2 to get x^{3}+2x^{2}-21x+98. Solve the equation where the result equals to 0.

$$x^{3}+2x^{2}-21x+98=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 98 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±98,±49,±14,±7,±2,±1$$

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.

$$x=-7$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}-21x+98 by x+7 to get x^{2}-5x+14. Solve the equation where the result equals to 0.

$$x^{2}-5x+14=0$$

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, -5 for b, and 14 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 14}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{-31}}{2}$$

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.

$$x=2 x=-7$$