Solve sqrt{25x^2-50}-sqrt{16x^2-32}=sqrt{13}

$$\sqrt { 25 x ^ { 2 } - 50 } - \sqrt { 16 x ^ { 2 } - 32 } = \sqrt { 13 }$$

Solve for x

$$x=\sqrt{15}\approx 3.872983346$$

$$x=-\sqrt{15}\approx -3.872983346$$

Solution Steps

Subtract -\sqrt{16x^{2}-32} from both sides of the equation.

$$\sqrt{25x^{2}-50}=\sqrt{13}+\sqrt{16x^{2}-32}$$

Square both sides of the equation.

$$\left(\sqrt{25x^{2}-50}\right)^{2}=\left(\sqrt{13}+\sqrt{16x^{2}-32}\right)^{2}$$

Calculate \sqrt{25x^{2}-50} to the power of 2 and get 25x^{2}-50.

$$25x^{2}-50=\left(\sqrt{13}+\sqrt{16x^{2}-32}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{13}+\sqrt{16x^{2}-32}\right)^{2}.

$$25x^{2}-50=\left(\sqrt{13}\right)^{2}+2\sqrt{13}\sqrt{16x^{2}-32}+\left(\sqrt{16x^{2}-32}\right)^{2}$$

The square of \sqrt{13} is 13.

$$25x^{2}-50=13+2\sqrt{13}\sqrt{16x^{2}-32}+\left(\sqrt{16x^{2}-32}\right)^{2}$$

Calculate \sqrt{16x^{2}-32} to the power of 2 and get 16x^{2}-32.

$$25x^{2}-50=13+2\sqrt{13}\sqrt{16x^{2}-32}+16x^{2}-32$$

Subtract 32 from 13 to get -19.

$$25x^{2}-50=-19+2\sqrt{13}\sqrt{16x^{2}-32}+16x^{2}$$

Subtract -19+16x^{2} from both sides of the equation.

$$25x^{2}-50-\left(-19+16x^{2}\right)=2\sqrt{13}\sqrt{16x^{2}-32}$$

To find the opposite of -19+16x^{2}, find the opposite of each term.

$$25x^{2}-50+19-16x^{2}=2\sqrt{13}\sqrt{16x^{2}-32}$$

Add -50 and 19 to get -31.

$$25x^{2}-31-16x^{2}=2\sqrt{13}\sqrt{16x^{2}-32}$$

Combine 25x^{2} and -16x^{2} to get 9x^{2}.

$$9x^{2}-31=2\sqrt{13}\sqrt{16x^{2}-32}$$

Square both sides of the equation.

$$\left(9x^{2}-31\right)^{2}=\left(2\sqrt{13}\sqrt{16x^{2}-32}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9x^{2}-31\right)^{2}.

$$81\left(x^{2}\right)^{2}-558x^{2}+961=\left(2\sqrt{13}\sqrt{16x^{2}-32}\right)^{2}$$

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

$$81x^{4}-558x^{2}+961=\left(2\sqrt{13}\sqrt{16x^{2}-32}\right)^{2}$$

Expand \left(2\sqrt{13}\sqrt{16x^{2}-32}\right)^{2}.

$$81x^{4}-558x^{2}+961=2^{2}\left(\sqrt{13}\right)^{2}\left(\sqrt{16x^{2}-32}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$81x^{4}-558x^{2}+961=4\left(\sqrt{13}\right)^{2}\left(\sqrt{16x^{2}-32}\right)^{2}$$

The square of \sqrt{13} is 13.

$$81x^{4}-558x^{2}+961=4\times 13\left(\sqrt{16x^{2}-32}\right)^{2}$$

Multiply 4 and 13 to get 52.

$$81x^{4}-558x^{2}+961=52\left(\sqrt{16x^{2}-32}\right)^{2}$$

Calculate \sqrt{16x^{2}-32} to the power of 2 and get 16x^{2}-32.

$$81x^{4}-558x^{2}+961=52\left(16x^{2}-32\right)$$

Use the distributive property to multiply 52 by 16x^{2}-32.

$$81x^{4}-558x^{2}+961=832x^{2}-1664$$

Subtract 832x^{2} from both sides.

$$81x^{4}-558x^{2}+961-832x^{2}=-1664$$

Combine -558x^{2} and -832x^{2} to get -1390x^{2}.

$$81x^{4}-1390x^{2}+961=-1664$$

Add 1664 to both sides.

$$81x^{4}-1390x^{2}+961+1664=0$$

Add 961 and 1664 to get 2625.

$$81x^{4}-1390x^{2}+2625=0$$

Substitute t for x^{2}.

$$81t^{2}-1390t+2625=0$$

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 81 for a, -1390 for b, and 2625 for c in the quadratic formula.

$$t=\frac{-\left(-1390\right)±\sqrt{\left(-1390\right)^{2}-4\times 81\times 2625}}{2\times 81}$$

Do the calculations.

$$t=\frac{1390±1040}{162}$$

Solve the equation t=\frac{1390±1040}{162} when ± is plus and when ± is minus.

$$t=15 t=\frac{175}{81}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=\sqrt{15} x=-\sqrt{15} x=\frac{5\sqrt{7}}{9} x=-\frac{5\sqrt{7}}{9}$$

Substitute \sqrt{15} for x in the equation \sqrt{25x^{2}-50}-\sqrt{16x^{2}-32}=\sqrt{13}.

$$\sqrt{25\left(\sqrt{15}\right)^{2}-50}-\sqrt{16\left(\sqrt{15}\right)^{2}-32}=\sqrt{13}$$

Simplify. The value x=\sqrt{15} satisfies the equation.

$$13^{\frac{1}{2}}=13^{\frac{1}{2}}$$

Substitute -\sqrt{15} for x in the equation \sqrt{25x^{2}-50}-\sqrt{16x^{2}-32}=\sqrt{13}.

$$\sqrt{25\left(-\sqrt{15}\right)^{2}-50}-\sqrt{16\left(-\sqrt{15}\right)^{2}-32}=\sqrt{13}$$

Simplify. The value x=-\sqrt{15} satisfies the equation.

$$13^{\frac{1}{2}}=13^{\frac{1}{2}}$$

Substitute \frac{5\sqrt{7}}{9} for x in the equation \sqrt{25x^{2}-50}-\sqrt{16x^{2}-32}=\sqrt{13}.

$$\sqrt{25\times \left(\frac{5\sqrt{7}}{9}\right)^{2}-50}-\sqrt{16\times \left(\frac{5\sqrt{7}}{9}\right)^{2}-32}=\sqrt{13}$$

Simplify. The value x=\frac{5\sqrt{7}}{9} does not satisfy the equation.

$$\frac{1}{9}\times 13^{\frac{1}{2}}=13^{\frac{1}{2}}$$

Substitute -\frac{5\sqrt{7}}{9} for x in the equation \sqrt{25x^{2}-50}-\sqrt{16x^{2}-32}=\sqrt{13}.

$$\sqrt{25\left(-\frac{5\sqrt{7}}{9}\right)^{2}-50}-\sqrt{16\left(-\frac{5\sqrt{7}}{9}\right)^{2}-32}=\sqrt{13}$$

Simplify. The value x=-\frac{5\sqrt{7}}{9} does not satisfy the equation.

$$\frac{1}{9}\times 13^{\frac{1}{2}}=13^{\frac{1}{2}}$$

Substitute \sqrt{15} for x in the equation \sqrt{25x^{2}-50}-\sqrt{16x^{2}-32}=\sqrt{13}.

$$\sqrt{25\left(\sqrt{15}\right)^{2}-50}-\sqrt{16\left(\sqrt{15}\right)^{2}-32}=\sqrt{13}$$

Simplify. The value x=\sqrt{15} satisfies the equation.

$$13^{\frac{1}{2}}=13^{\frac{1}{2}}$$

Substitute -\sqrt{15} for x in the equation \sqrt{25x^{2}-50}-\sqrt{16x^{2}-32}=\sqrt{13}.

$$\sqrt{25\left(-\sqrt{15}\right)^{2}-50}-\sqrt{16\left(-\sqrt{15}\right)^{2}-32}=\sqrt{13}$$

Simplify. The value x=-\sqrt{15} satisfies the equation.

$$13^{\frac{1}{2}}=13^{\frac{1}{2}}$$

List all solutions of \sqrt{25x^{2}-50}=\sqrt{16x^{2}-32}+\sqrt{13}.

$$x=\sqrt{15} x=-\sqrt{15}$$