In this video lesson, I review the basic concepts of percentages and percent changes that are tested on the SAT math section.

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In this video lesson, I review the basic concepts of percentages and percent changes that are tested on the SAT math section.

In this video lesson, I review the basic concepts of fractional exponents that are tested on the SAT math section.

In this video lesson, I review the basics of the general equation of a parabola that shows up in different forms on the SAT math.

**Key Concept**: If you are given the equation of a circle in the expanded form $$ x^2 – 2hx + y^2 – 2ky = f $$ then the center of the circle is $(h, k)$ and the radius is given by $r = \sqrt{f+h^2+k^2}$.

The above result saves the effort of completing the square and turning the above equation into the general equation of a circle. We will apply this concept to solve a difficult SAT math problem.

**Practice Problem**: The equation of a circle in the $xy$-plane is shown below. What is the radius of the circle?

$$ 2x^2 – 2x + 2y^2 + 10y = 5 $$

- $\sqrt{5}$
- $3$
- $4$
- $5$

**Solution**: First divide the equation by 2

$$ 2x^2 – 2x + 2y^2 + 10y = 5 $$

and rewrite in a form where the coefficients of $x^2$ and $y^2$ terms are equal to 1. We obtain

$$x^2 – x + y^2 + 5y = \dfrac{5}{2}$$

If we compare this to the general expanded form for the equation of a circle

$$ x^2 – 2hx + y^2 – 2ky = f$$

we obtain $-2h=-1$ and $-2k=5$, therefore $h=\dfrac{1}{2}$ and $k=-\dfrac{5}{2}$, and the center of the circle is

$\left(\dfrac{1}{2}, -\dfrac{5}{2}\right)$.

Also, $f=\dfrac{5}{2}$, and we know that $r = \sqrt{f+h^2+k^2}$, where $f=\dfrac{5}{2}$, $h=\dfrac{1}{2}$, and $k=-\dfrac{5}{2}$. On substitution, we obtain $$r = \sqrt{\dfrac{5}{2} + \left(\dfrac{1}{2}\right)^2 + \left(-\dfrac{5}{2}\right)^2} = \sqrt{\dfrac{5}{2} + \dfrac{1}{4} + \dfrac{25}{4}} = \sqrt{\dfrac{10+1+25}{4}} = \sqrt{\dfrac{36}{4}}=\sqrt{9}=3$$

**Explanation**: The result stated above can be obtained by starting with the general equation of a circle with center at $(h, k)$ and a radius of $r$ units:

$$

\begin{align}

(x-h)^2 + (y-k)^2 & = r^2 && \text{Square the algebraic terms}\\

x^2 – 2hx + h^2 + y^2 -2ky + k^2 & = r^2 && \text{Move the constant terms to the right} \\

x^2 – 2hx + y^2 -2ky & = r^2 – h^2 – k^2 && \\

\end{align}

$$

Compare the above equation to the general expanded form

$$ x^2 – 2hx + y^2 – 2ky = f $$

and we find that the coordinates of the center $(h, k)$ appear as part of the coefficients of the $x$ and $y$ terms. Further, $f = r^2 – h^2 – k^2$, which on rearrangment gives $r = \sqrt{f+h^2+k^2}$.

A quadratic equation has the form $ax^2 + bx + c=0$, where $a, b$, and $c$ are real numbers, and $a \neq 0$. The $ax^2$ term is called the quadratic term, the $bx$ is the linear term, and $c$ is called the constant term. Consider the quadratic equation $x^2 – x -6=0$, if we substitute $x=3$ in this equation we find that the equation holds true because $3^2-3-6=9-9=0$. Similarly if we substitute $x=-2$ in this equation we again find that $ (-2)^2 – (-2) – 6 = 4+2-6=0$. These two values of $x$ are the only two numbers that satisfy this quadratic equation, and are called the roots of the quadratic equation. Sometimes the roots are also referred to as zeros of the quadratic equation, in other words the value of the quadratic expression is zero at these values.

The roots of the quadratic equation $x^2 – x -6=0$ can be obtained by factoring the equation into a product of binomials. Let’s assume that the binomials are $x+r_1$ and $x+r_2$, our goal is to find the values of $r_1$ and $r_2$ such that: $$(x+r_1)(x+r_2) = x^2 – x -6$$ $$x^2 + (r_1+r_2)x + r_1r_2 = x^2 – x -6$$ The two expressions above are identical and therefore the coefficient of $x$ terms must be equal or $r_1+r_2=-1$. Similarly, the constant terms must also be equal, therefore $r_1 r_2=-6$. By using trial and error we find that the values of $r_1$ and $r_2$ are $2$ and $-3$. Therefore our quadratic equation can be rewritten as $$x^2 – x -6 = (x+2)(x-3) =0$$ For the above equation to hold true, either $x+2=0$ or $x-3=0$, which tells us that $x=-2$ and $x=3$ are the solutions to this quadratic equation.

When the coefficient of the $x^2$ term is not 1 we follow the steps described below to factor the quadratic equation. I will use the example of $2x^2-5x-33=0$.

- Multiply the coefficient of $x^2$, $2$ in this case, and the constant term of $-33$, which gives a value of $-66$.
- Use trial and error to find two numbers that multiply to give $-66$ but add up to the coefficient of $x$, which is $-5$.
- The two numbers are $-11$ and $6$.
- Rewrite the middle term as the sum of these two numbers $-5x = -11x+6x$.

$$ 2x^2-5x-33=2x^2-11x + 6x-33=0 $$ Factor the largest common term from the first two terms, and also from the last two terms. $$ x(2x-11) + 3(2x-11)=0$$ $$(2x-11)(x + 3)=0 \quad \text{Factor $(2x-11)$} $$ - The roots are then obtained by solving $2x-11=0$ and $x+3=0$, which gives $\dfrac{11}{2}$ and $-3$ as the

two roots of the quadratic equation $2x^2-5x-33=0$.

Completing a square refers to the process of adding a constant term to a quadratic expression such that the result is a perfect square. Consider the expression $x^2+6x+4$, we can rewrite this as $$x^2+6x+9-9+4=x^2+6x+9-5$$ We further recognize that $x^2+6x+9=(x+3)^2$ is a perfect square, and therefore the original expression is equivalent to $(x+3)^2-5$.

In general if we are given a quadratic of the form $x^2+bx$, we add $\left(\dfrac{b}{2}\right)^2$ to the quadratic to obtain a perfect square. $$ x^2 + bx + \left(\dfrac{b}{2}\right)^2 = x^2+ bx + \dfrac{b^2}{4} = \left(x+ \dfrac{b}{2}\right)^2$$

Completing the square of a quadratic expression helps us to find the least and greatest value of such expressions. In general the quadratic expression $ax^2+bx+c$ attains its maxima ($a<0$) or minima ($a>0$) when $x = -\dfrac{b}{2a}$. The greatest and the least value of the quadratic expression can be found by substituting $x=-\dfrac{b}{2a}$ into the expression $ax^2+bx+c$.

**Solved Example:** $$ h(t) = -16t^2+24t+18$$

The equation above gives the height above ground $h$, in feet, of a ball $t$ seconds after it is launched vertically upward from a platform that is at a height of 18 feet. What is the greatest height the ball achieves?

- $21$
- $24$
- $27$
- $30$

**Explanation:**The easiest way to solve this problem is by recognizing that the maximum value of the expression occurs when $t = -\dfrac{b}{2a}= -\dfrac{24}{2(-16)} = \dfrac{3}{4}$, and we can find the maximum height by substituting $t=\dfrac{3}{4}$ in the expression $-16t^2+24t+18= -16\left(\dfrac{3}{4}\right)^2 + 24\left(\dfrac{3}{4}\right) + 18 = -9 + 18 + 18 = 27$.

We can also do this problem by completing the square, we first factor out the term $-16$ from the first two terms of the quadratic expression to obtain:

$$

\begin{align}

h(t) &= -16t^2+24t+18 && \text{Factor out $-16$}\\

&= -16\left[t^2 – \dfrac{3}{2}\right] + 18 && \text{Complete the square} \\

&=-16\left[t^2 – \dfrac{3}{2}+ \left(\dfrac{3}{4}\right)^2 – \left(\dfrac{3}{4}\right)^2\right] + 18 && \text{} \\

&= -16\left[t^2 – \dfrac{3}{2}+ \left(\dfrac{3}{4}\right)^2\right] +16\left(\dfrac{3}{4}\right)^2+ 18 && \text{}\\

& = -16\left(t-\dfrac{3}{4}\right)^2 + 27 && \\

\end{align}

$$

Note that the expression $\left(t-\dfrac{3}{4}\right)^2 \geq 0$ because it is a square of a number. This means that the term $-16\left(t-\dfrac{3}{4}\right)^2$ will always be less than or equal to zero. The maximum height will have a value of $27$ and this will happen when $t=\dfrac{3}{4}$.

The quadratic formula can be used to obtain the solutions to the general quadratic equation $ax^2 + bx + c=0$:

$$ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $$

**Example**: $2x^2-5x-33=0$, we have $a=2$, $b=-5$, and $c=-33$, replacing these values in the above quadratic formula yields $$ x = \dfrac{-(-5) \pm \sqrt{(-5)^2-4(2)(-33)}}{2(2)}=\dfrac{5 \pm \sqrt{289}}{4} = \dfrac{5 \pm 17}{4}$$ and the two solutions are $x=\dfrac{11}{2}$ and $x=-3$.

**Solved Example:** $$ \dfrac{x^2}{2} = mx + \dfrac{n}{2} $$

In the quadratic equation above, $m$ and $n$ are constants. What are the solutions for $x$ ?

- $\quad m \pm \sqrt{m^2-n}$
- $\quad m \pm \sqrt{m^2+n}$
- $\quad -m \pm \sqrt{m^2-n}$
- $\quad -m \pm \sqrt{m^2+n}$

**Explanation:** $$

\begin{align}

\dfrac{x^2}{2} &= mx + \dfrac{n}{2} && \text{Multiply both sides by 2}\\

x^2 &= 2mx +n && \text{Rewrite in standard form} \\

x^2 – 2mx – n &=0 && \text{Note $a=1$, $b=-2m$, $c=-n$} \\

x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} && \text{Apply quadratic formula}\\

& = \dfrac{-(-2m) \pm \sqrt{(-2m)^2-4(1)(-n)}}{2(1)} && \\

& = \dfrac{2m \pm \sqrt{4m^2+4n}}{2} &&\\

& = \dfrac{2m \pm 2\sqrt{m^2+n}}{2} && \text{Divide throughout by 2}\\

& = m \pm \sqrt{m^2+n} &&\\

\end{align}

$$

Let $r_1$ and $r_2$ be the roots of the quadratic equation $ax^2 + bx + c=0$, then the quadratic equation in the factored form can be written as $a(x-r_1)(x-r_2)=0$. If we multiply this expression and rearrange it, we obtain:

$$

\begin{align}

a(x-r_1)(x-r_2)&=0 &&\\

a(x^2 – xr_2 – xr_1 + r_1 r_2) &=0 &&\\

a[x^2 -x(r_1+r_2)+r_1r_2] &=0 &&\\

ax^2-xa(r_1+r_2)+ar_1r_2 &=0 &&\\

\end{align}

$$

This resulting expanded form is equivalent to the quadratic equation $ax^2 + bx + c=0$, and the coefficient of all the terms must be identical. If we equate the coefficient of $x$ term we obtain $-a(r_1+r_2) = b$, or in other words:

$$\textrm{Sum of Roots} = r_1 + r_2 = -\dfrac{b}{a} $$ Further if we equate the constant part of the two equations, we obtain $ar_1r_2=c$, or in other words: $$ \textrm{Product of Roots} = r_1 r_2 = \frac{c}{a} $$ The above two formulas are helpful in finding the sum and product of the roots without having to find the roots themselves.

**Solved Example:** $$ 2x^2=3x+2 $$ What is the sum of all the solutions to the quadratic equation above?

- $\quad -\dfrac{5}{2}$
- $\quad -\dfrac{3}{2}$
- $\quad \quad \dfrac{3}{2}$
- $\quad \quad \dfrac{5}{2}$

**Explanation:** Rewrite the equation in the standard quadratic form $ax^2+bx+c=0$ as $2x^2-3x-2=0$. We have $a=2$, $b=-3$, and $c=-2$. The sum of the roots is equal to $-\dfrac{b}{a}=-\dfrac{(-3)}{2} = \dfrac{3}{2}$. There is no need to waste time in first finding the two roots, which are $-\dfrac{1}{2}$ and $2$, and adding them to find their sum of $\dfrac{3}{2}$.

The table below summarizes how the discriminant $b^2-4ac$ determines the nature of the roots of a general quadratic equation $ax^2+bx+c=0$, and where the corresponding graph of the parabola $y=ax^2+bx+c=0$ falls with respect to the $x-$axis. I will review parabolas in a different article.

$$

\begin{array}{|c|c|c|}

\hline

\textbf{Condition} & \textbf{Nature of Roots} & \textbf{Graph of the}\\

& & \textbf{corresponding parabola} \\ \hline

b^2 – 4ac > 0 & \text{Roots are distinct and real} & \text{Parabola intersects the $x-$axis}\\

& & \text{at two distinct points}\\ \hline

b^2 – 4ac < 0 & \text{No real roots} & \text{Parabola does not }\\
& & \text{intersect the $x-$axis}\\ \hline
b^2 - 4ac = 0 & \text{Roots are identical} & \text{Parabola is tangent} \\
& \text{also called a double root} & \text{to the $x-$axis}\\ \hline
\end{array}
$$
**Example:** If $b$ is a constant such that the equation $2x^2=bx-3$ has two distinct real solutions, which of the following could be the value of $b$ ?

- $\quad -5$
- $\quad -4$
- $\quad -3$
- $\quad 4$

**Explanation:** First write the quadratic equation in the standard form $2x^2-bx+3=0$. For the roots to be real and distinct $b^2-4ac>0$, which is equivalent to $(-b)^2 – 4(2)(3)>0$ or $b^2>24$. The only value in the answer choices that satisfies this condition is $-5$.