Solve (7x+1)^2=3(x)

$$( 7 x + 1 ) ^ { 2 } = 3 ( x )$$

Solve for x (complex solution)

$$x=\frac{-11+5\sqrt{3}i}{98}\approx -0.112244898+0.088369939i$$

$$x=\frac{-5\sqrt{3}i-11}{98}\approx -0.112244898-0.088369939i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7x+1\right)^{2}.

$$49x^{2}+14x+1=3x$$

Subtract 3x from both sides.

$$49x^{2}+14x+1-3x=0$$

Combine 14x and -3x to get 11x.

$$49x^{2}+11x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, 11 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 49}}{2\times 49}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 49}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-11±\sqrt{121-196}}{2\times 49}$$

Add 121 to -196.

$$x=\frac{-11±\sqrt{-75}}{2\times 49}$$

Take the square root of -75.

$$x=\frac{-11±5\sqrt{3}i}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{-11±5\sqrt{3}i}{98}$$

Now solve the equation x=\frac{-11±5\sqrt{3}i}{98} when ± is plus. Add -11 to 5i\sqrt{3}.

$$x=\frac{-11+5\sqrt{3}i}{98}$$

Now solve the equation x=\frac{-11±5\sqrt{3}i}{98} when ± is minus. Subtract 5i\sqrt{3} from -11.

$$x=\frac{-5\sqrt{3}i-11}{98}$$

The equation is now solved.

$$x=\frac{-11+5\sqrt{3}i}{98} x=\frac{-5\sqrt{3}i-11}{98}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7x+1\right)^{2}.

$$49x^{2}+14x+1=3x$$

Subtract 3x from both sides.

$$49x^{2}+14x+1-3x=0$$

Combine 14x and -3x to get 11x.

$$49x^{2}+11x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$49x^{2}+11x=-1$$

Divide both sides by 49.

$$\frac{49x^{2}+11x}{49}=-\frac{1}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$x^{2}+\frac{11}{49}x=-\frac{1}{49}$$

Divide \frac{11}{49}, the coefficient of the x term, by 2 to get \frac{11}{98}. Then add the square of \frac{11}{98} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{49}x+\left(\frac{11}{98}\right)^{2}=-\frac{1}{49}+\left(\frac{11}{98}\right)^{2}$$

Square \frac{11}{98} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{49}x+\frac{121}{9604}=-\frac{1}{49}+\frac{121}{9604}$$

Add -\frac{1}{49} to \frac{121}{9604} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{49}x+\frac{121}{9604}=-\frac{75}{9604}$$

Factor x^{2}+\frac{11}{49}x+\frac{121}{9604}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{98}\right)^{2}=-\frac{75}{9604}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{98}\right)^{2}}=\sqrt{-\frac{75}{9604}}$$

Simplify.

$$x+\frac{11}{98}=\frac{5\sqrt{3}i}{98} x+\frac{11}{98}=-\frac{5\sqrt{3}i}{98}$$

Subtract \frac{11}{98} from both sides of the equation.

$$x=\frac{-11+5\sqrt{3}i}{98} x=\frac{-5\sqrt{3}i-11}{98}$$