4x+6/12x+4=2x/6-এর সমাধান করুন

$$\frac { 4 x + 6 } { 12 x + 4 } = \frac { 2 x } { 6 }$$

Solve for x

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3}\approx 1.602628851$$

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3}\approx -0.935962184$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 12\left(3x+1\right), the least common multiple of 12x+4,6.

$$3\left(4x+6\right)=\left(6x+2\right)\times 2x$$

Use the distributive property to multiply 3 by 4x+6.

$$12x+18=\left(6x+2\right)\times 2x$$

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

$$12x+18=\left(12x+4\right)x$$

Use the distributive property to multiply 12x+4 by x.

$$12x+18=12x^{2}+4x$$

Subtract 12x^{2} from both sides.

$$12x+18-12x^{2}=4x$$

Subtract 4x from both sides.

$$12x+18-12x^{2}-4x=0$$

Combine 12x and -4x to get 8x.

$$8x+18-12x^{2}=0$$

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.

$$-12x^{2}+8x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 8 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-12\right)\times 18}}{2\left(-12\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-12\right)\times 18}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-8±\sqrt{64+48\times 18}}{2\left(-12\right)}$$

Multiply 48 times 18.

$$x=\frac{-8±\sqrt{64+864}}{2\left(-12\right)}$$

Add 64 to 864.

$$x=\frac{-8±\sqrt{928}}{2\left(-12\right)}$$

Take the square root of 928.

$$x=\frac{-8±4\sqrt{58}}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-8±4\sqrt{58}}{-24}$$

Now solve the equation x=\frac{-8±4\sqrt{58}}{-24} when ± is plus. Add -8 to 4\sqrt{58}.

$$x=\frac{4\sqrt{58}-8}{-24}$$

Divide -8+4\sqrt{58} by -24.

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Now solve the equation x=\frac{-8±4\sqrt{58}}{-24} when ± is minus. Subtract 4\sqrt{58} from -8.

$$x=\frac{-4\sqrt{58}-8}{-24}$$

Divide -8-4\sqrt{58} by -24.

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3}$$

The equation is now solved.

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3} x=\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 12\left(3x+1\right), the least common multiple of 12x+4,6.

$$3\left(4x+6\right)=\left(6x+2\right)\times 2x$$

Use the distributive property to multiply 3 by 4x+6.

$$12x+18=\left(6x+2\right)\times 2x$$

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

$$12x+18=\left(12x+4\right)x$$

Use the distributive property to multiply 12x+4 by x.

$$12x+18=12x^{2}+4x$$

Subtract 12x^{2} from both sides.

$$12x+18-12x^{2}=4x$$

Subtract 4x from both sides.

$$12x+18-12x^{2}-4x=0$$

Combine 12x and -4x to get 8x.

$$8x+18-12x^{2}=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$8x-12x^{2}=-18$$

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.

$$-12x^{2}+8x=-18$$

Divide both sides by -12.

$$\frac{-12x^{2}+8x}{-12}=-\frac{18}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{8}{-12}x=-\frac{18}{-12}$$

Reduce the fraction \frac{8}{-12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{2}{3}x=-\frac{18}{-12}$$

Reduce the fraction \frac{-18}{-12} to lowest terms by extracting and canceling out 6.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{3}{2}+\frac{1}{9}$$

Add \frac{3}{2} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{29}{18}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{29}{18}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{29}{18}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{58}}{6} x-\frac{1}{3}=-\frac{\sqrt{58}}{6}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3} x=-\frac{\sqrt{58}}{6}+\frac{1}{3}$$