Solve (sqrt{3}-5x)^2-5(x-sqrt{3})^2

$$( \sqrt { 3 } - 5 x ) ^ { 2 } - 5 ( x - \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$20x^{2}-12$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-5x\right)^{2}.

$$\left(\sqrt{3}\right)^{2}-10\sqrt{3}x+25x^{2}-5\left(x-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$3-10\sqrt{3}x+25x^{2}-5\left(x-\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\sqrt{3}\right)^{2}.

$$3-10\sqrt{3}x+25x^{2}-5\left(x^{2}-2x\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$3-10\sqrt{3}x+25x^{2}-5\left(x^{2}-2x\sqrt{3}+3\right)$$

Use the distributive property to multiply -5 by x^{2}-2x\sqrt{3}+3.

$$3-10\sqrt{3}x+25x^{2}-5x^{2}+10\sqrt{3}x-15$$

Combine 25x^{2} and -5x^{2} to get 20x^{2}.

$$3-10\sqrt{3}x+20x^{2}+10\sqrt{3}x-15$$

Combine -10\sqrt{3}x and 10\sqrt{3}x to get 0.

$$3+20x^{2}-15$$

Subtract 15 from 3 to get -12.

$$-12+20x^{2}$$

Expand

$$20x^{2}-12$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-5x\right)^{2}.

$$\left(\sqrt{3}\right)^{2}-10\sqrt{3}x+25x^{2}-5\left(x-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$3-10\sqrt{3}x+25x^{2}-5\left(x-\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\sqrt{3}\right)^{2}.

$$3-10\sqrt{3}x+25x^{2}-5\left(x^{2}-2x\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$3-10\sqrt{3}x+25x^{2}-5\left(x^{2}-2x\sqrt{3}+3\right)$$

Use the distributive property to multiply -5 by x^{2}-2x\sqrt{3}+3.

$$3-10\sqrt{3}x+25x^{2}-5x^{2}+10\sqrt{3}x-15$$

Combine 25x^{2} and -5x^{2} to get 20x^{2}.

$$3-10\sqrt{3}x+20x^{2}+10\sqrt{3}x-15$$

Combine -10\sqrt{3}x and 10\sqrt{3}x to get 0.

$$3+20x^{2}-15$$

Subtract 15 from 3 to get -12.

$$-12+20x^{2}$$