### Solving Equations without a Linear Term

Quadratic equations with no linear term can be solved by isolating $x^2$ and taking the square root of both sides of the equation. If $x^2$ is equal to a negative number, then the solutions of the equation are pure imaginary numbers.

A
Solving an equation of the form $ax^2 = -c$ combines the processes of solving a quadratic equation and simplifying imaginary numbers.

**quadratic equation**is an equation, where $a$, $b$, and $c$ are real numbers and $a\neq0$, written in the form:$ax^2 + bx + c = 0$

Step-By-Step Example

Solving a Quadratic Equation of the Form $ax^2 = -c$

Solve the equation:

$5x^2=-80$

Step 1

Divide both sides of the equation by the coefficient of $x^2$.

$\begin{aligned}\frac {5x^2}{5}&=\frac{-80}{5}\\x^2&=-16\end{aligned}$

Step 2

Take the square root of both sides of the equation.

$\begin{aligned}\sqrt{(x^2)}&=\sqrt{-16}\\x&=\pm \sqrt{-16}\end{aligned}$

Solution

Simplify the negative square root.
The solutions are $x=4i$ or $x=-4i$.

$\begin{aligned}x&=\pm \sqrt{-16}\\&=\pm \sqrt{-1\cdot 16}\\&=\pm (\sqrt{-1}\cdot \sqrt{16})\\&=\pm (i\cdot4)\\&=\pm 4i\end{aligned}$

### Solving by Completing the Square

Quadratic equations with no real roots can be solved by completing the square.

The roots of a quadratic equation are its solutions. A **real root** is a solution of an equation that is also a real number. Some quadratic equations do not have any real roots.

$f(x)=ax^2+bx+c$

When a quadratic equation has no real roots, it has a pair of complex roots that are complex conjugates. One strategy for solving quadratic equations with complex roots is completing the square.

Completing the square is the process of adding a constant term to a quadratic expression to form a perfect square trinomial. A **perfect square trinomial** is a trinomial, or polynomial with three terms, that can be written as the square of a binomial. For example, $x^2+8x+16$ is a perfect square trinomial because its factored form is $(x+4)^2$.

$\left(\frac{b}{2}\right)^{2}$

Step-By-Step Example

Complete the Square to Solve a Quadratic Equation with Complex Solutions

Solve the equation:

$x^2-6x+10=0$

Step 1

Write the equation in the form $x^2+bx=c$.

$x^2-6x=-10$

Step 2

Find the value of this formula:

$\begin{aligned}\left (\frac{b}{2} \right )^{2}&=\left ( \frac{-6}{2} \right )^{2}\\&=(-3)^{2}\\&=9\end{aligned}$

Step 3

Add 9 to $x^2+bx$ to complete the square. Add 9 to the other side of the equation to keep the equation balanced.

$\begin{aligned}x^2-6x+9&=-10+9\\x^2-6x+9&=-1\end{aligned}$

Step 4

Factor the perfect square trinomial, or a polynomial with three terms that can be written as the square of a binomial.

$\left(x-3\right)^2=-1$

Step 5

Take the square root of both sides of the equation.

$\begin{aligned}x-3&=\pm \sqrt{-1}\\x-3&=\pm i\end{aligned}$

Solution

Use both the positive and negative square root to write and solve two equations.
The solutions are:
The solutions can be checked by substituting them into the original equation.
For $x=3+i$:
For $x=3-i$:

$\begin{aligned}x-3&=i\\x&=3+i\end{aligned}\;\;\;\;\; \text{or}\;\;\;\;\;\begin{aligned}x-3&=-i\\x&=3-i\end{aligned}$

$x=3+i\;\;\;\;\; \text{or}\;\;\;\;\;x=3-i$

$\begin{aligned}x^2-6x+10&=0\\(3+i)^2-6(3+i)+10&=0\\(3+i)(3+i)-6(3+i)+10&=0\\9+3i+3i+i^2-18-6i+10&=0\\1+(-1)&=0\\0&=0\end{aligned}$

$\begin{aligned}x^2-6x+10&=0\\(3-i)^2-6(3-i)+10&=0\\(3-i)(3-i)-6(3-i)+10&=0\\9-3i-3i+i^2-18+6i+10&=0\\1+(-1)&=0\\0&=0\end{aligned}$

### Solving by Using the Quadratic Formula

Quadratic equations with no real roots can be solved by using the quadratic formula.

Another way to find solutions to quadratic equations with no real roots is by using the quadratic formula. The
The quadratic formula can be used to find the solutions of
The value of the discriminant indicates the type and number of roots of the quadratic equation.

**quadratic formula**is a formula used to solve a quadratic equation of the form $ax^2+bx+c=0$ by using its coefficients:$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

*any*quadratic equation. Once the equation is in the standard form of a quadratic equation, $ax^2+bx+c=0$, substitute the values of $a$, $b$, and $c$ into the formula and simplify. The**discriminant**is the part of the quadratic formula that is under the radical sign:$b^2-4ac$

- If the discriminant is positive, then the quadratic equation has two real roots.
- If the discriminant is zero, then the quadratic equation has one real root.
- If the discriminant is negative, then the quadratic equation has two complex roots, which come in conjugate pairs.

Step-By-Step Example

Using the Quadratic Formula with Complex Solutions

Solve the equation:

$4x^2-16x+41=0$

Step 1

The standard form of an equation is:
The given equation is in standard form, where $a=4$, $b=-16$, and $c=41$. Substitute the values of $a$, $b$, and $c$ into the discriminant from the quadratic formula.
The discriminant is negative. So the equation has two complex solutions.

$ax^2+bx+c=0$

$\begin{aligned}b^2-4ac&=(-16)^2-4(4)(41)\\&=-400\end{aligned}$

Step 2

Substitute the values of $a$, $b$, and the discriminant into the quadratic formula.

$\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-(-16)\pm\sqrt{-400}}{2(4)}\end{aligned}$

Solution

Simplify the expression.
The solutions are:

$\begin{aligned}x&=\frac{16\pm\sqrt{-400}}{8}\\&=\frac{16\pm 20i}{8}\\&=2\pm\frac{5}{2}i\end{aligned}$

$x=2+\frac{5}{2}i\;\;\;\;\;\text{or}\;\;\;\;\;x=2-\frac{5}{2}i$