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A quadratic equation is an equation where the highest degree is $2.$ Quadratic equations coincide with quadratic functions, and can be used to solve for various points on the graph, or parabola, of the function. There are many different ways to solve quadratic equations. One way is with the use of square roots.

In standard form, quadratic equations take the form $y=ax^2+bx+c,$ and can be solved in various ways. In general, they are solved for the value(s) of $x$ that make the equation equal to $0.$ Thus, $y=ax^2+bx+c$ becomes $0=ax^2+bx+c.$ Graphically, all points with a $y$-coordinate of $0$ are the $x$-intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have $0, 1,$ or $2$ zeros, a quadratic equation can have $0, 1,$ or $2$ solutions.

Simple quadratic equations take the form $ax^2+c=0$ and are solved using inverse operations. Once $x^2$ remains on the left-hand side, the equation can be written as $x^2=d,$ where $d=\frac{\text{-} c}{a}$. The value of $d$ gives the number of solutions the equation has.

$\begin{aligned} \mathbf{d>0} \quad &: \quad 2\text{ real solutions}\\ \mathbf{d=0} \quad &: \quad 1\text{ real solution}\\ \mathbf{d<0} \quad &: \quad 0\text{ real solutions} \end{aligned}$Without solving completely, determine the number of solutions the quadratic equation has. $x^2+75=50$

Show Solution

A quadratic equation can have $0, 1,$ or $2$ real solutions. Without solving completely, it's possible to determine how many just by isolating $x^2.$ Here, this means subtracting $75$ on both sides of the equation.

$x^2+75=50 \quad \Rightarrow \quad x^2=\text{-} 25$ Since $\text{-} 25 <0,$ the equation has $0$ real solutions. Since the solutions to a quadratic equation give the zeros of the parabola, we can conclude that the parabola does not intersect the $x$-axis.

Quadratic equations of the form
$ax^2+c=0$
can be solved using inverse operations. To undo the exponent of $2$ on the $x^2$ term, square roots must be used. Consider $5x^2-500=0.$
### 1

To begin, isolate $x^2$ using inverse operations to move $c$ and then $a$ to the opposite side of the equation. Here, that means adding $500$ and then dividing by $5.$
### 2

Now that $x^2$ has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Thus, square roots undo powers of $2$ and cube roots undo powers of $3,$ etc. To finish isolating $x^2,$ square-root both sides of the equation.
Using a calculator to determine $\sqrt{100}$ gives one value — $10.$ This is because calculators give the principal root of a square root. It is necessary to remember that there are two values that, when raised to the power of $2,$ equal $100.$
$10 \cdot 10 = 100 \quad \text{and} \quad (\text{-} 10)(\text{-} 10)=100,$
Thus, $x=10$ and $x=\text{-} 10.$

Isolate $x^2$

Square-root both sides of the equation

The area of a square measures $81 \text{ cm}^2.$ Write a simple quadratic equation to represent the area. Then, determine the side length of the square.

Show Solution

To begin, recall that the area of a square can be found using the formula
$A=s^2,$
where $s$ is the length of one of the sides. We can write an equation to represent the area of the given square by substituting $A=81.$ This gives $81=s^2.$
Solving this equation for $s$ gives the side's length. Since $s$ is raised to the power of $2,$ we can square root both sides.
The values of $s$ that satisfy the created equation are $s=9$ and $s=\text{-} 9.$ Since $s$ represents a side length, it can only be positive. Thus, $s=9$ is the only viable solution. The side length of the square measures $9$ cm.

$81 = s^2$

SqrtEqn$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$\pm \sqrt{81}= s$

CalcRootCalculate root

$\pm 9 = s$

RearrangeEqnRearrange equation

$s = \pm 9$

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