Equation Lab #15 – Quadratic Equations: Square Roots

It’s that time of the month again, so here’s another week of Equation Lab lessons. During the last two months, I went over how to solve linear equations, as standalone equations and in systems. By now, we are done with solving linear equations, so it’s time to move up to the next level. Our first one is the next basic equation – quadratic equations.

Introduction:

A quadratic equation is an equation where the term with the highest exponent has an exponent of 2. Here are some examples of quadratic equations:

• $x^{2}+1=5$
• $2x^{2}=18$
• $x^{2}+6x+8=0$

Notice that they all have an $x^{2}$ term. That’s what determines if they’re a quadratic equation. If they have an $x^{2}$ term, and no other term with a higher exponent, then they are quadratic equations.

The definition of a quadradic equation is:

• $y=ax^{2}+bx+c$

Where $a$ is the coefficient of the quadratic term, $b$ is the coefficient of the linear term, and $c$ is the constant.

Quadratic equations can have two real solutions, one real solution, or no real solution. The number of solutions is determined by the discriminant, which follows the formula $b^{2}-4ac$. If the discriminant is positive, there will be two solutions. If the discriminant is zero, there will only be one solution. And if the discriminant is negative, then in terms of real numbers, there are no solutions.

There are five ways to solve quadratic equations. And this time, the methods don’t work on all types. Here are the methods and when to use them:

• Graphing – if you have a graphing calculator or graph paper, you can graph the parabola. The x-intercepts (the points where $y=0$) of a parabola are the solutions. If there are two x-intercepts, there are two solutions. If there is one x-intercept, there’s only one solution. If there are no x-intercepts, there are no solutions.
• Square Roots – the square roots method is an additional step to solving single-variable equations if you’re used to the four basic operators. Like linear equations, you’ll have to isolate the variable to solve the equation. When you’re down to the base quadratic variable, you can take the square root of both sides. Your result is the square root and its negative equivalent.
  • This method only applies to quadratic equations where $b=0$. In other words, there must be no linear terms in order to use this method.
• Factoring – the factoring method involves factoring the quadratic trinomial. The equation must be set to equal $0$. Then you ca;n factor the trinomial into two linear binomials, where each one could equal $0$. You solve for each binomial to gather the solution.
  • You can use this method on both equations with a linear term and equations without a linear term, but you better watch out. If the calculated value for $b^{2}-4ac$ is not a perfect square, then the trinomial is irreducible, and thus cannot be factored.
• Completing the Square – the completing the square method involves algebraic manipulation where you turn the left side of the equation into a perfect square trinomial. The right side should be a single constant, while the left side should be a perfect square trinomial. Once you get there, you can take the square root of both sides, then solve the linear equation.
  • This method only applies to quadratic equations where $b\ne 0$. In other words, there must be a linear term in order to use this method.
• Quadratic Formula – the ultimate way to solve a quadratic equation. This method can be used to solve any quadratic equation. It involves a formula where you evaluate the constants and coefficients.

Other than the graphing method, all of these methods will be covered this week.

The Square Roots Method:

Starting with the easiest type of quadratic equation, the steps to solving a linear equation is the same as solving a quadratic equation of this method. Just like how you previously had to isolate the variable to solve a linear equation, you’ll have to isolate the quadratic term. That means, it must have to constants or coefficients applied. Once it’s down to the quadratic term, you can take the square root. In general, you will not just have a solution, but rather two solutions. Remember that the constant on the right side of the equation must be positive for this to happen.

Let’s start with the first equation, $\frac{1}{3}m^{2}+8=20$. What numbers are applied to the $m^{2}$ term? The answers are $8$ and $3$. How are they applied? By addition and division, respectively. So, to solve for $m^{2}$, you must subtract $8$ and multiply by $3$. By subtracting $8$ from both sides, the equation becomes $\frac{1}{3}m^{2}=12$. By multiplying both sides by $3$, the equation becomes $m^{2}=36$.

Now for the big step. This is when you’ll have to take the square root of both sides. $\sqrt{m^{2}}=m$, as$\sqrt{36}=6$. But we aren’t taking $\sqrt{36}$. We’re taking $\pm\sqrt{36}$. The result is $\pm 6$. This means that $m=6$ or $m=-6$.

1. $\frac{1}{3}m^{2}+8=20$
2. $\frac{1}{3}m^{2}+8-8=20-8$
3. $\frac{1}{3}m^{2}=12$
4. $\frac{1}{3}m^{2}\cdot 3=12\cdot 3$
5. $m^2=36$
6. $\sqrt{m^{2}}=\pm\sqrt{36}$
7. $m=6, m=-6$

Checking our work, let’s plug in each value for $m$. Whether $m=6$ or $m=-6$, $m^{2}=36$ since both $6\cdot 6=36$ and $-6\cdot-6=36$. But $36/3=12$. Add $8$ to $12$, you’ll get $20$. So $m=6$ or $m=-6$ when $\frac{1}{3}m^{2}+8=20$.

Let’s do another example. The equation is $5p^{2}-10=70$. What numbers are applied to the $p^{2}$ term? The answers are $10$ and $5$. How are they applied? By subtraction and multiplication, respectively. So, to solve for $p^{2}$, you must add $10$ and divide by $5$. By adding $10$ to both sides, the equation becomes $5p^{2}=80$. By dividing both sides by $5$, the equation becomes $p^{2}=16$.

All what’s left is to take the square root of both sides. The square root of $16$ is $4$, but since we’re counting both negative and positive solutions, $p=4$ or $p=-4$.

1. $5p^{2}-10=70$
2. $5p^{2}-10+10=70+10$
3. $5p^{2}=80$
4. $5p^{2}/5=80/5$
5. $p^{2}=16$
6. $\sqrt{p^{2}}=\pm\sqrt{16}$
7. $p=4, p=-4$

Checking our work, let’s plug in each value for $p$. Whether $p=4$ or $p=-4$, $p^{2}=16$ since both $4\cdot 4=16$ and $-4 \cdot -4=16$. But $16\cdot 5=80$, and $80-10=70$. So $p=4$ or $p=-4$ when $5p^{2}-10=70$.

And that’s how the square root method works. Let’s do our first word problem involving quadratic equations.

Word Problem #13:

Bug is presenting a present to Guy for his birthday in a box of a perfect cube. The box containing Bug’s present for Guy has a surface area of 1,350 square inches. What is the volume of the box?

Let’s define our variable. According to the geometric formulas, which I’ll cover later this year, the equation for the surface area of a cube is $SA=6s^{2}$. So our variable is $s$. We are already given the surface area, which is $1350$. So the equation to solve is $6s^{2}=1350$.

With no constant applied to the equation, we can simply divide both sides by $6$. $1350/6=225$, so the equation becomes $s^{2}=225$. Take the square root of both sides, you should get $s=\pm15$ since the square root of $225$ is $15$. However, when solving real-world problems like this, we do not take negative numbers into consideration since it’s not possible to have a negative amount of something. Therefore, the sole solution is $s=15$.

The final step is to raise $15$ to an exponent of $3$ since we’re looking for the volume, which is defined as $V=s^{3}$ for cubes. $15^{3}=3375$.

 1. Since the present is a perfect cube,
    use $SA=6s^{2}$.
 2. The present has a surface area of
    1,350 square inches.
 3. $6s^{2}=1350$
 4. $6s^{2}/6=1350/6$
 5. $s^{2}=225$
 6. $\sqrt{s^{2}}=\sqrt{225}$
 7. $s=15$
 8. $V=s^{3}$
 9. $V=(15)^{3}$
10. $V=3375$

The box has a volume of 3,375 cubic inches.

Quiz #15:

Now that the lesson is over, let’s see if you can complete this quiz. Like always, the correct solutions and the correct variables must be used. But for this quiz, I will take any answer for as long as it fits the solution. So if the answer is $a=3,-3$, you can use “$3$“, “$-3$“, “$3,-3″$, “$-3,3$“, “$a=3$“, “$a=-3$“, “$a=3,-3$“, or “$a=-3,3$“.