Solve 0^2=(25x^2-60x-36)(25x^2+60x+36)

$$0 ^ { 2 } = ( 25 x ^ { 2 } - 60 x - 36 ) ( 25 x ^ { 2 } + 60 x + 36 )$$

Solve for x

$$x = \frac{6 \sqrt{2} + 6}{5} \approx 2.897056275$$

$$x=\frac{6-6\sqrt{2}}{5}\approx -0.497056275$$

$$x = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

Solution Steps

Calculate 0 to the power of 2 and get 0.

$$0=\left(25x^{2}-60x-36\right)\left(25x^{2}+60x+36\right)$$

Use the distributive property to multiply 25x^{2}-60x-36 by 25x^{2}+60x+36 and combine like terms.

$$0=625x^{4}-3600x^{2}-4320x-1296$$

Swap sides so that all variable terms are on the left hand side.

$$625x^{4}-3600x^{2}-4320x-1296=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1296 and q divides the leading coefficient 625. List all candidates \frac{p}{q}.

$$±\frac{1296}{625},±\frac{1296}{125},±\frac{1296}{25},±\frac{1296}{5},±1296,±\frac{648}{625},±\frac{648}{125},±\frac{648}{25},±\frac{648}{5},±648,±\frac{432}{625},±\frac{432}{125},±\frac{432}{25},±\frac{432}{5},±432,±\frac{324}{625},±\frac{324}{125},±\frac{324}{25},±\frac{324}{5},±324,±\frac{216}{625},±\frac{216}{125},±\frac{216}{25},±\frac{216}{5},±216,±\frac{162}{625},±\frac{162}{125},±\frac{162}{25},±\frac{162}{5},±162,±\frac{144}{625},±\frac{144}{125},±\frac{144}{25},±\frac{144}{5},±144,±\frac{108}{625},±\frac{108}{125},±\frac{108}{25},±\frac{108}{5},±108,±\frac{81}{625},±\frac{81}{125},±\frac{81}{25},±\frac{81}{5},±81,±\frac{72}{625},±\frac{72}{125},±\frac{72}{25},±\frac{72}{5},±72,±\frac{54}{625},±\frac{54}{125},±\frac{54}{25},±\frac{54}{5},±54,±\frac{48}{625},±\frac{48}{125},±\frac{48}{25},±\frac{48}{5},±48,±\frac{36}{625},±\frac{36}{125},±\frac{36}{25},±\frac{36}{5},±36,±\frac{27}{625},±\frac{27}{125},±\frac{27}{25},±\frac{27}{5},±27,±\frac{24}{625},±\frac{24}{125},±\frac{24}{25},±\frac{24}{5},±24,±\frac{18}{625},±\frac{18}{125},±\frac{18}{25},±\frac{18}{5},±18,±\frac{16}{625},±\frac{16}{125},±\frac{16}{25},±\frac{16}{5},±16,±\frac{12}{625},±\frac{12}{125},±\frac{12}{25},±\frac{12}{5},±12,±\frac{9}{625},±\frac{9}{125},±\frac{9}{25},±\frac{9}{5},±9,±\frac{8}{625},±\frac{8}{125},±\frac{8}{25},±\frac{8}{5},±8,±\frac{6}{625},±\frac{6}{125},±\frac{6}{25},±\frac{6}{5},±6,±\frac{4}{625},±\frac{4}{125},±\frac{4}{25},±\frac{4}{5},±4,±\frac{3}{625},±\frac{3}{125},±\frac{3}{25},±\frac{3}{5},±3,±\frac{2}{625},±\frac{2}{125},±\frac{2}{25},±\frac{2}{5},±2,±\frac{1}{625},±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

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.

$$x=-\frac{6}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 625x^{4}-3600x^{2}-4320x-1296 by 5\left(x+\frac{6}{5}\right)=5x+6 to get 125x^{3}-150x^{2}-540x-216. Solve the equation where the result equals to 0.

$$125x^{3}-150x^{2}-540x-216=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -216 and q divides the leading coefficient 125. List all candidates \frac{p}{q}.

$$±\frac{216}{125},±\frac{216}{25},±\frac{216}{5},±216,±\frac{108}{125},±\frac{108}{25},±\frac{108}{5},±108,±\frac{72}{125},±\frac{72}{25},±\frac{72}{5},±72,±\frac{54}{125},±\frac{54}{25},±\frac{54}{5},±54,±\frac{36}{125},±\frac{36}{25},±\frac{36}{5},±36,±\frac{27}{125},±\frac{27}{25},±\frac{27}{5},±27,±\frac{24}{125},±\frac{24}{25},±\frac{24}{5},±24,±\frac{18}{125},±\frac{18}{25},±\frac{18}{5},±18,±\frac{12}{125},±\frac{12}{25},±\frac{12}{5},±12,±\frac{9}{125},±\frac{9}{25},±\frac{9}{5},±9,±\frac{8}{125},±\frac{8}{25},±\frac{8}{5},±8,±\frac{6}{125},±\frac{6}{25},±\frac{6}{5},±6,±\frac{4}{125},±\frac{4}{25},±\frac{4}{5},±4,±\frac{3}{125},±\frac{3}{25},±\frac{3}{5},±3,±\frac{2}{125},±\frac{2}{25},±\frac{2}{5},±2,±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

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.

$$x=-\frac{6}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 125x^{3}-150x^{2}-540x-216 by 5\left(x+\frac{6}{5}\right)=5x+6 to get 25x^{2}-60x-36. Solve the equation where the result equals to 0.

$$25x^{2}-60x-36=0$$

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 25 for a, -60 for b, and -36 for c in the quadratic formula.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 25\left(-36\right)}}{2\times 25}$$

Do the calculations.

$$x=\frac{60±60\sqrt{2}}{50}$$

Solve the equation 25x^{2}-60x-36=0 when ± is plus and when ± is minus.

$$x=\frac{6-6\sqrt{2}}{5} x=\frac{6\sqrt{2}+6}{5}$$

List all found solutions.

$$x=-\frac{6}{5} x=\frac{6-6\sqrt{2}}{5} x=\frac{6\sqrt{2}+6}{5}$$