Solve 25x=sqrt{49x^2+48^2}

$$25x= \sqrt{ 49 { x }^{ 2 } + { 48 }^{ 2 } }$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(25x\right)^{2}=\left(\sqrt{49x^{2}+48^{2}}\right)^{2}$$

Expand \left(25x\right)^{2}.

$$25^{2}x^{2}=\left(\sqrt{49x^{2}+48^{2}}\right)^{2}$$

Calculate 25 to the power of 2 and get 625.

$$625x^{2}=\left(\sqrt{49x^{2}+48^{2}}\right)^{2}$$

Calculate 48 to the power of 2 and get 2304.

$$625x^{2}=\left(\sqrt{49x^{2}+2304}\right)^{2}$$

Calculate \sqrt{49x^{2}+2304} to the power of 2 and get 49x^{2}+2304.

$$625x^{2}=49x^{2}+2304$$

Subtract 49x^{2} from both sides.

$$625x^{2}-49x^{2}=2304$$

Combine 625x^{2} and -49x^{2} to get 576x^{2}.

$$576x^{2}=2304$$

Subtract 2304 from both sides.

$$576x^{2}-2304=0$$

Divide both sides by 576.

$$x^{2}-4=0$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Substitute 2 for x in the equation 25x=\sqrt{49x^{2}+48^{2}}.

$$25\times 2=\sqrt{49\times 2^{2}+48^{2}}$$

Simplify. The value x=2 satisfies the equation.

$$50=50$$

Substitute -2 for x in the equation 25x=\sqrt{49x^{2}+48^{2}}.

$$25\left(-2\right)=\sqrt{49\left(-2\right)^{2}+48^{2}}$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-50=50$$

Equation 25x=\sqrt{49x^{2}+2304} has a unique solution.

$$x=2$$