# How to Solve a Quadratic With Imaginary Numbers

By Kylene Arnold
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The solutions to a quadratic equation are defined as the two x-intercepts of its graph, or the two values of x that cause the value of y to equal zero. However, certain quadratics have no x-intercepts, which means that using only real numbers, y cannot equal zero. One might think that these equations are unsolvable. However, this is not so. These contrary equations are quite solvable using the imaginary number, i, a theoretical number equal to the square root of -1. Using imaginary numbers, these "unsolvable" quadratics can be solved using the same two methods used to solve conventional quadratics: completing the square and the quadratic formula.

## Completing the Square

Make sure the equation is the in the standard form, ax^2 + bx + c = 0, where "a" is the coefficient of x^2, "b" is the coefficient of x and "c" is a real number. If the equation is not in standard form, use algebra to rearrange the equation. For example, 2x^2 + 4x = -6 is not in standard form. Using algebra, you can add 6 to both sides of the equation to get 2x^2 + 4x + 6 = 0. The equation is now in standard form.

Note that "a" must always be positive. If it is not, multiply the whole equation by -1.

Make sure that "a" equals 1. If it does not, multiply the whole equation by whatever number will make "a" equal 1. For instance, in the equation 2x^2 + 4x + 6 = 0, "a" equals 2. Multiply the whole equation by 1/2 to make "a" equal 1:

(1/2)(2x^2 + 4x + 6 = 0) = x^2 + 2x + 3 = 0.

Subtract "c" from both sides of the equation. In our equation x^2 + 2x + 3 = 0, "c" equals 3. Subtract 3 from both sides to get x^2 + 2x = -3

Divide "b" by 2. Square the result and add it to both sides. In our sample equation x^2 + 2x = -3, "b" equals 2, and 2 divided by 2 equals 1. 1 squared is still 1. Add 1 to both sides of the equation to get x^2 + 2x + 1 = -2.

Note that in equations that require imaginary numbers, the right hand side will always be negative at the end of this step.

Simplify the left hand side of the equation into a perfect square. A perfect square is a binomial (an equation with two parts, or terms) that when squared results in a trinomial (an equation with three terms). In this example, x^2 + 2x + 1 simplifies to (x + 1)^2. The equation now reads (x + 1)^2 = -2.

Use your calculator to take the square root of both sides. To take the square root of the negative number on the right hand side, remember that algebra allows a single root to also be expressed as the product of two other roots. Therefore, the root of a negative number can be expressed as the product of the root of the corresponding positive number times the square root of -1, which is the imaginary number, i. Also, do not forget that there is both a positive and a negative root to all real numbers. You will need to notate this in your answer with the following symbol: ±.

In our example, the equation √(x + 1)^2 = √-2 becomes x + 1= ± 1.4i.

The square root of 2 is a non-terminating decimal, so it has been rounded to the nearest tenths place.

Add or subtract from both sides to solve for x. Do not forget that there is both a positive and a negative number on the right hand side, as indicated by the "±" symbol. Add or subtract from both numbers to get the two solutions to the quadratic.

In our example, x = ±1.4i - 1, which means that x = 1.4i - 1 and x = -1.4i - 1. Therefore, x = 0.4i and -2.4i

Make sure the equation is the in the standard form, ax^2 + bx + c = 0, where "a" is the coefficient of x^2, "b" is the coefficient of x and "c" is a real number. If the equation is not in standard form, use algebra to rearrange the equation.

For example, 2x^2 + 4x = -6 is not in standard form. Using algebra, you can add 6 to both sides of the equation to get 2x^2 + 4x + 6 = 0. The equation is now in standard form.

Substitute the values of a, b and c in the equation into the quadratic formula: [-b ± √(b^2 - 4ac)]/ 2a = x.

For example, substituting the values from the equation 2x^2 + 4x + 6 = 0 into the quadratic formula, you get {-4 ± √[4^2 - 4(2)(6)]}/ 2(2) = x.

Remember to include the signs that go with the coefficients when substituting numbers into the quadratic equation. For example, if the equation had read 2x^2 - 4x - 6, b would have been -4 and c, -6.

Simplify the denominator and the expression under the radical (the square root sign). Your equation should now look like this:

(-4 ± √-32)/ 4 = x.

Because the equation requires you to take the square root of a negative number, you know that you will need imaginary numbers to calculate the solution.

Use your calculator to take the square root of the negative number. To do this, remember that algebra allows a single root to also be expressed as the product of two other roots. Therefore, the root of a negative number can be expressed as the product of the root of the corresponding positive number times the square root of -1, which is the imaginary number, i.

In our example, √-32 = i(√32), which equals 5.7i. The whole equation now reads like this: (-4 ± 5.7i)/4 = x.

The square root of 32 is a non-terminating decimal and has been rounded to the nearest tenth.

Break the complex fraction on the left hand side of the equation into the sum of its component fractions. In this case, (-4 ± 5.7i)/4 can be broken up into the sum of its two simpler fractions:

-4/4 ± 5.7i/4.

Simplify the fractions by dividing the numerator by the denominator. In this example, -4/4 ± 5.7i/4 simplifies into -1 ± 1.4i. The whole equation now reads -1 ± 1.4i = x.

All decimals have been rounded to the nearest tenth.

Solve for x. Do not forget you have both a positive and a negative complex number, as indicated by the "±" symbol. (A complex number is the number that was formed from the product of a real number and the imaginary number, i.) Add or subtract from both the positive and the negative number to get the two solutions to the quadratic.

In this example, -1 ± 1.4i = x, which means that -1 + 1.4i = x and -1 - 1.4i = x. Therefore, x = 0.4i and -2.4i.