Solve 62/x+77/x+6=3

$$\frac { 62 } { x } + \frac { 77 } { x + 6 } = 3$$

Solve for x

$$x = \frac{\sqrt{19105} + 121}{6} \approx 43.203472965$$

$$x=\frac{121-\sqrt{19105}}{6}\approx -2.870139632$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+6\right), the least common multiple of x,x+6.

$$\left(x+6\right)\times 62+x\times 77=3x\left(x+6\right)$$

Use the distributive property to multiply x+6 by 62.

$$62x+372+x\times 77=3x\left(x+6\right)$$

Combine 62x and x\times 77 to get 139x.

$$139x+372=3x\left(x+6\right)$$

Use the distributive property to multiply 3x by x+6.

$$139x+372=3x^{2}+18x$$

Subtract 3x^{2} from both sides.

$$139x+372-3x^{2}=18x$$

Subtract 18x from both sides.

$$139x+372-3x^{2}-18x=0$$

Combine 139x and -18x to get 121x.

$$121x+372-3x^{2}=0$$

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.

$$-3x^{2}+121x+372=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 121 for b, and 372 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-121±\sqrt{121^{2}-4\left(-3\right)\times 372}}{2\left(-3\right)}$$

Square 121.

$$x=\frac{-121±\sqrt{14641-4\left(-3\right)\times 372}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-121±\sqrt{14641+12\times 372}}{2\left(-3\right)}$$

Multiply 12 times 372.

$$x=\frac{-121±\sqrt{14641+4464}}{2\left(-3\right)}$$

Add 14641 to 4464.

$$x=\frac{-121±\sqrt{19105}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-121±\sqrt{19105}}{-6}$$

Now solve the equation x=\frac{-121±\sqrt{19105}}{-6} when ± is plus. Add -121 to \sqrt{19105}.

$$x=\frac{\sqrt{19105}-121}{-6}$$

Divide -121+\sqrt{19105} by -6.

$$x=\frac{121-\sqrt{19105}}{6}$$

Now solve the equation x=\frac{-121±\sqrt{19105}}{-6} when ± is minus. Subtract \sqrt{19105} from -121.

$$x=\frac{-\sqrt{19105}-121}{-6}$$

Divide -121-\sqrt{19105} by -6.

$$x=\frac{\sqrt{19105}+121}{6}$$

The equation is now solved.

$$x=\frac{121-\sqrt{19105}}{6} x=\frac{\sqrt{19105}+121}{6}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+6\right), the least common multiple of x,x+6.

$$\left(x+6\right)\times 62+x\times 77=3x\left(x+6\right)$$

Use the distributive property to multiply x+6 by 62.

$$62x+372+x\times 77=3x\left(x+6\right)$$

Combine 62x and x\times 77 to get 139x.

$$139x+372=3x\left(x+6\right)$$

Use the distributive property to multiply 3x by x+6.

$$139x+372=3x^{2}+18x$$

Subtract 3x^{2} from both sides.

$$139x+372-3x^{2}=18x$$

Subtract 18x from both sides.

$$139x+372-3x^{2}-18x=0$$

Combine 139x and -18x to get 121x.

$$121x+372-3x^{2}=0$$

Subtract 372 from both sides. Anything subtracted from zero gives its negation.

$$121x-3x^{2}=-372$$

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.

$$-3x^{2}+121x=-372$$

Divide both sides by -3.

$$\frac{-3x^{2}+121x}{-3}=-\frac{372}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{121}{-3}x=-\frac{372}{-3}$$

Divide 121 by -3.

$$x^{2}-\frac{121}{3}x=-\frac{372}{-3}$$

Divide -372 by -3.

$$x^{2}-\frac{121}{3}x=124$$

Divide -\frac{121}{3}, the coefficient of the x term, by 2 to get -\frac{121}{6}. Then add the square of -\frac{121}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{121}{3}x+\left(-\frac{121}{6}\right)^{2}=124+\left(-\frac{121}{6}\right)^{2}$$

Square -\frac{121}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{121}{3}x+\frac{14641}{36}=124+\frac{14641}{36}$$

Add 124 to \frac{14641}{36}.

$$x^{2}-\frac{121}{3}x+\frac{14641}{36}=\frac{19105}{36}$$

Factor x^{2}-\frac{121}{3}x+\frac{14641}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{121}{6}\right)^{2}=\frac{19105}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{121}{6}\right)^{2}}=\sqrt{\frac{19105}{36}}$$

Simplify.

$$x-\frac{121}{6}=\frac{\sqrt{19105}}{6} x-\frac{121}{6}=-\frac{\sqrt{19105}}{6}$$

Add \frac{121}{6} to both sides of the equation.

$$x=\frac{\sqrt{19105}+121}{6} x=\frac{121-\sqrt{19105}}{6}$$