Solve {left(sqrt{(x+14)*3x}right)}^2=x+14

$${ \left( \sqrt{ (x+14) \times 3x } \right) }^{ 2 } =x+14$$

Solve for x

$$x=-14$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+14$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+14$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+14$$

Subtract x from both sides.

$$3x^{2}+42x-x=14$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=14$$

Subtract 14 from both sides.

$$3x^{2}+41x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=41 ab=3\left(-14\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 41.

$$a=-1 b=42$$

Rewrite 3x^{2}+41x-14 as \left(3x^{2}-x\right)+\left(42x-14\right).

$$\left(3x^{2}-x\right)+\left(42x-14\right)$$

Factor out x in the first and 14 in the second group.

$$x\left(3x-1\right)+14\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(x+14\right)$$

To find equation solutions, solve 3x-1=0 and x+14=0.

$$x=\frac{1}{3} x=-14$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+14$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+14$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+14$$

Subtract x from both sides.

$$3x^{2}+42x-x=14$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=14$$

Subtract 14 from both sides.

$$3x^{2}+41x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 41 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}-4\times 3\left(-14\right)}}{2\times 3}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\times 3\left(-14\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-41±\sqrt{1681-12\left(-14\right)}}{2\times 3}$$

Multiply -12 times -14.

$$x=\frac{-41±\sqrt{1681+168}}{2\times 3}$$

Add 1681 to 168.

$$x=\frac{-41±\sqrt{1849}}{2\times 3}$$

Take the square root of 1849.

$$x=\frac{-41±43}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-41±43}{6}$$

Now solve the equation x=\frac{-41±43}{6} when ± is plus. Add -41 to 43.

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

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-41±43}{6} when ± is minus. Subtract 43 from -41.

$$x=-\frac{84}{6}$$

Divide -84 by 6.

$$x=-14$$

The equation is now solved.

$$x=\frac{1}{3} x=-14$$

Steps for Completing the Square

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+14$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+14$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+14$$

Subtract x from both sides.

$$3x^{2}+42x-x=14$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=14$$

Divide both sides by 3.

$$\frac{3x^{2}+41x}{3}=\frac{14}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{41}{3}x=\frac{14}{3}$$

Divide \frac{41}{3}, the coefficient of the x term, by 2 to get \frac{41}{6}. Then add the square of \frac{41}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{41}{3}x+\left(\frac{41}{6}\right)^{2}=\frac{14}{3}+\left(\frac{41}{6}\right)^{2}$$

Square \frac{41}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{41}{3}x+\frac{1681}{36}=\frac{14}{3}+\frac{1681}{36}$$

Add \frac{14}{3} to \frac{1681}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{41}{3}x+\frac{1681}{36}=\frac{1849}{36}$$

Factor x^{2}+\frac{41}{3}x+\frac{1681}{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}.

$$\left(x+\frac{41}{6}\right)^{2}=\frac{1849}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{41}{6}\right)^{2}}=\sqrt{\frac{1849}{36}}$$

Simplify.

$$x+\frac{41}{6}=\frac{43}{6} x+\frac{41}{6}=-\frac{43}{6}$$

Subtract \frac{41}{6} from both sides of the equation.

$$x=\frac{1}{3} x=-14$$