Solve 2x+3/x-2=1/4quadx=?

$$\frac { 2 x + 3 } { x - 2 } = \frac { 1 } { 4 } \quad x = ?$$

Solve for x

$$x=\sqrt{37}+5\approx 11.08276253$$

$$x=5-\sqrt{37}\approx -1.08276253$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-2\right), the least common multiple of x-2,4.

$$4\left(2x+3\right)=\frac{1}{4}x\times 4\left(x-2\right)$$

Use the distributive property to multiply 4 by 2x+3.

$$8x+12=\frac{1}{4}x\times 4\left(x-2\right)$$

Multiply \frac{1}{4} and 4 to get 1.

$$8x+12=x\left(x-2\right)$$

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

$$8x+12=x^{2}-2x$$

Subtract x^{2} from both sides.

$$8x+12-x^{2}=-2x$$

Add 2x to both sides.

$$8x+12-x^{2}+2x=0$$

Combine 8x and 2x to get 10x.

$$10x+12-x^{2}=0$$

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.

$$-x^{2}+10x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-10±\sqrt{100+48}}{2\left(-1\right)}$$

Add 100 to 48.

$$x=\frac{-10±\sqrt{148}}{2\left(-1\right)}$$

Take the square root of 148.

$$x=\frac{-10±2\sqrt{37}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±2\sqrt{37}}{-2}$$

Now solve the equation x=\frac{-10±2\sqrt{37}}{-2} when ± is plus. Add -10 to 2\sqrt{37}.

$$x=\frac{2\sqrt{37}-10}{-2}$$

Divide -10+2\sqrt{37} by -2.

$$x=5-\sqrt{37}$$

Now solve the equation x=\frac{-10±2\sqrt{37}}{-2} when ± is minus. Subtract 2\sqrt{37} from -10.

$$x=\frac{-2\sqrt{37}-10}{-2}$$

Divide -10-2\sqrt{37} by -2.

$$x=\sqrt{37}+5$$

The equation is now solved.

$$x=5-\sqrt{37} x=\sqrt{37}+5$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-2\right), the least common multiple of x-2,4.

$$4\left(2x+3\right)=\frac{1}{4}x\times 4\left(x-2\right)$$

Use the distributive property to multiply 4 by 2x+3.

$$8x+12=\frac{1}{4}x\times 4\left(x-2\right)$$

Multiply \frac{1}{4} and 4 to get 1.

$$8x+12=x\left(x-2\right)$$

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

$$8x+12=x^{2}-2x$$

Subtract x^{2} from both sides.

$$8x+12-x^{2}=-2x$$

Add 2x to both sides.

$$8x+12-x^{2}+2x=0$$

Combine 8x and 2x to get 10x.

$$10x+12-x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$10x-x^{2}=-12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+10x=-12$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=-\frac{12}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=-\frac{12}{-1}$$

Divide -12 by -1.

$$x^{2}-10x=12$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=12+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=12+25$$

Add 12 to 25.

$$x^{2}-10x+25=37$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=37$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{37}$$

Simplify.

$$x-5=\sqrt{37} x-5=-\sqrt{37}$$

Add 5 to both sides of the equation.

$$x=\sqrt{37}+5 x=5-\sqrt{37}$$