Solve (x+sqrt{3})(x-sqrt{3})=6(x-2)

$$(x+ \sqrt{ 3 } )(x- \sqrt{ 3 } )=6(x-2)$$

Solve for x

$$x=3$$

Steps Using the Quadratic Formula

Consider \left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-\left(\sqrt{3}\right)^{2}=6\left(x-2\right)$$

The square of \sqrt{3} is 3.

$$x^{2}-3=6\left(x-2\right)$$

Use the distributive property to multiply 6 by x-2.

$$x^{2}-3=6x-12$$

Subtract 6x from both sides.

$$x^{2}-3-6x=-12$$

Add 12 to both sides.

$$x^{2}-3-6x+12=0$$

Add -3 and 12 to get 9.

$$x^{2}+9-6x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}-6x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2}$$

Add 36 to -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-6}{2}$$

The opposite of -6 is 6.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Steps for Completing the Square

Consider \left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-\left(\sqrt{3}\right)^{2}=6\left(x-2\right)$$

The square of \sqrt{3} is 3.

$$x^{2}-3=6\left(x-2\right)$$

Use the distributive property to multiply 6 by x-2.

$$x^{2}-3=6x-12$$

Subtract 6x from both sides.

$$x^{2}-3-6x=-12$$

Add 3 to both sides.

$$x^{2}-6x=-12+3$$

Add -12 and 3 to get -9.

$$x^{2}-6x=-9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$