ACT Math 3: Intermediate Algebra — Quadratics and Polynomials

Quadratic equation

An equation containing a second-degree polynomial, usually written in standard form as $ax^2+bx+c=0$, where $a$, $b$, and $c$ are real numbers and $a\ne0$.

What is the zero-product property, and why is it useful for solving quadratics?

If $AB=0$, then $A=0$ or $B=0$. After factoring a quadratic into a product of factors equal to zero, set each factor equal to zero to find the solutions.

What conditions must be met before solving a quadratic by factoring?

Rewrite the equation in standard form, $ax^2+bx+c=0$, with zero on one side. Then factor the entire expression, including any greatest common factor, and apply the zero-product property.

How can $x^2+bx+c=0$ be factored when the leading coefficient is $1$?

Find numbers $m$ and $n$ such that $mn=c$ and $m+n=b$. Then write $(x+m)(x+n)=0$ and solve each linear factor.

Solve $x^2+x-6=0$ by factoring.

$x^2+x-6=(x-2)(x+3)$, so $x=2$ or $x=-3$.

Difference of squares

A binomial of the form $u^2-v^2$ factors as $(u-v)(u+v)$. For example, $x^2-9=(x-3)(x+3)$.

Solve $x^2-25=0$ using factoring.

$x^2-25=(x-5)(x+5)$, so the solutions are $x=5$ and $x=-5$.

How is a quadratic with leading coefficient $a\ne1$ commonly factored by grouping?

For $ax^2+bx+c$, calculate $ac$, find two numbers whose product is $ac$ and sum is $b$, split the $bx$ term using those numbers, and factor by grouping.

Solve $4x^2+15x+9=0$ by factoring by grouping.

Since $4\cdot9=36$ and $3+12=15$, rewrite the expression as $4x^2+3x+12x+9$. It factors as $(4x+3)(x+3)=0$, giving $x=-\frac{3}{4}$ or $x=-3$.

Why should the greatest common factor be checked first when solving a polynomial equation?

Factoring out the GCF can reveal a factor such as $x$ and reduce the remaining expression to a factorable quadratic. Each resulting factor can then produce a solution through the zero-product property.

Solve $-3x^3-5x^2-2x=0$ by factoring.

Factor to obtain $-x(3x+2)(x+1)=0$. The solutions are $x=0$, $x=-\frac{2}{3}$, and $x=-1$.

Square root property

If $u^2=k$, then $u=\pm\sqrt{k}$. The $\pm$ is essential because both the positive and negative square roots have the same square.

When is the square root property especially efficient for solving a quadratic?

Use it when the equation has no linear term after rearrangement, such as $x^2=k$, or when a squared binomial is isolated, such as $(x-h)^2=k$.

What happens when the square root property gives $u^2=k$ with $k<0$?

There are no real solutions because the square of a real number cannot be negative. Over the complex numbers, the solutions are $u=\pm i\sqrt{-k}$.

Solve $4x^2+1=7$ using the square root property.

Subtract $1$ and divide by $4$: $x^2=\frac{3}{2}$. Therefore, $x=\pm\sqrt{\frac{3}{2}}=\pm\frac{\sqrt6}{2}$.

Solve $3(x-4)^2=15$ using the square root property.

Divide by $3$ to get $(x-4)^2=5$. Thus $x-4=\pm\sqrt5$, so $x=4\pm\sqrt5$.

Completing the square

A method that adds a carefully chosen constant to both sides so that a quadratic trinomial becomes a perfect-square binomial. For $x^2+bx$, add $\left(\frac{b}{2}\right)^2$.

What is the standard procedure for completing the square in $ax^2+bx+c=0$?

First divide by $a$ if necessary, move the constant term, and add $\left(\frac{b}{2}\right)^2$ to both sides. Factor the perfect square and use the square root property.

Solve $x^2+4x+1=0$ by completing the square.

Rewrite as $x^2+4x=-1$, then add $4$ to both sides: $(x+2)^2=3$. Therefore, $x=-2\pm\sqrt3$.

Solve $x^2-3x-5=0$ by completing the square.

Rearrange to $x^2-3x=5$ and add $\frac{9}{4}$: $(x-\frac{3}{2})^2=\frac{29}{4}$. Thus, $x=\frac{3\pm\sqrt{29}}{2}$.

Quadratic formula

For $ax^2+bx+c=0$ with $a\ne0$, the solutions are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. It can solve every quadratic equation.

What precautions reduce errors when using the quadratic formula?

Put the equation in standard form first, identify $a$, $b$, and $c$, and substitute negative coefficients in parentheses. Ensure that the entire numerator is divided by $2a$.

Solve $x^2+5x+1=0$ using the quadratic formula.

Here $a=1$, $b=5$, and $c=1$. The solutions are $x=\frac{-5\pm\sqrt{21}}{2}$.

What is the discriminant of a quadratic equation?

The discriminant is $\Delta=b^2-4ac$, the expression under the square root in the quadratic formula. Its value predicts the number and type of solutions.

What does each sign of the discriminant indicate for a quadratic with real coefficients?

If $\Delta>0$, there are two distinct real solutions; if $\Delta=0$, there is one repeated real solution; if $\Delta<0$, there are two nonreal complex-conjugate solutions.

How does a positive discriminant distinguish rational from irrational solutions?

When $\Delta>0$ is a perfect square, the two real solutions are rational. If $\Delta>0$ is not a perfect square, the two real solutions are irrational.

Determine the nature of the solutions to $3x^2-10x+15=0$ without solving.

$\Delta=(-10)^2-4(3)(15)=-80<0$. Therefore, the equation has two complex, nonreal solutions.

What is the graphical meaning of the solutions of $ax^2+bx+c=0$?

They are the $x$-intercepts, or zeros, of the parabola $y=ax^2+bx+c$. If the discriminant is negative, the graph has no real $x$-intercepts.

How can a quadratic equation be solved graphically when it is written as $f(x)=k$?

Graph $y=f(x)$ and the horizontal line $y=k$. The $x$-coordinates of their intersection points are the solutions.

What is the relationship between the quadratic formula and completing the square?

The quadratic formula is derived by applying the completing-the-square method to the general equation $ax^2+bx+c=0$. Thus, the formula is a general shortcut for that procedure.

What is the sum of the two solutions of $ax^2+bx+c=0$?

By the relationships between roots and coefficients, the sum is $x_1+x_2=-\frac{b}{a}$.

What is the product of the two solutions of $ax^2+bx+c=0$?

The product of the roots is $x_1x_2=\frac{c}{a}$.

A rectangular garden is $10$ feet longer than its width and has area $119\text{ ft}^2$. What equation models its dimensions?

If the width is $w$, the length is $w+10$, so $w(w+10)=119$, or $w^2+10w-119=0$. Factoring gives $w=7$ or $w=-17$; the valid dimensions are width $7$ ft and length $17$ ft.

How should extraneous algebraic roots be handled in a physical application of a quadratic?

Check each mathematical solution against the context. Reject values that violate restrictions such as negative lengths, negative time, or a specified interval.

A quantity follows $P=-t^2+13t+130$ for $1\le t\le6$. How would you find when $P=160$?

Set $-t^2+13t+130=160$, giving $t^2-13t+30=0=(t-3)(t-10)$. The algebraic solutions are $3$ and $10$, but the domain restriction leaves $t=3$ days.