Solve (x^2-5x)(x^2-5x)=196

$$( x ^ { 2 } - 5 x ) ( x ^ { 2 } - 5 x ) = 196$$

Solve for x (complex solution)

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

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

$$x=-2$$

$$x=7$$

Solution Steps

Multiply x^{2}-5x and x^{2}-5x to get \left(x^{2}-5x\right)^{2}.

$$\left(x^{2}-5x\right)^{2}=196$$

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

$$\left(x^{2}\right)^{2}-10x^{2}x+25x^{2}=196$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}-10x^{2}x+25x^{2}=196$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{4}-10x^{3}+25x^{2}=196$$

Subtract 196 from both sides.

$$x^{4}-10x^{3}+25x^{2}-196=0$$

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}-10x^{3}+25x^{2}-196 by x+2 to get x^{3}-12x^{2}+49x-98. Solve the equation where the result equals to 0.

$$x^{3}-12x^{2}+49x-98=0$$

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}-12x^{2}+49x-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

$$x=-2$$

$$x=7$$

Solution Steps

Multiply x^{2}-5x and x^{2}-5x to get \left(x^{2}-5x\right)^{2}.

$$\left(x^{2}-5x\right)^{2}=196$$

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

$$\left(x^{2}\right)^{2}-10x^{2}x+25x^{2}=196$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}-10x^{2}x+25x^{2}=196$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{4}-10x^{3}+25x^{2}=196$$

Subtract 196 from both sides.

$$x^{4}-10x^{3}+25x^{2}-196=0$$

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}-10x^{3}+25x^{2}-196 by x+2 to get x^{3}-12x^{2}+49x-98. Solve the equation where the result equals to 0.

$$x^{3}-12x^{2}+49x-98=0$$

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}-12x^{2}+49x-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.

$$x=-2 x=7$$