SAT Math 6: Nonlinear Equations in One Variable

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 standard form required for applying the quadratic formula?

The equation must be written as $ax^2+bx+c=0$, with $a\ne0$. The coefficients are then identified as $a$, $b$, and $c$.

Zero-product property

If a product equals zero, then at least one factor must equal zero: if $AB=0$, then $A=0$ or $B=0$. This property allows a factored quadratic to be solved by setting each factor equal to zero.

Why must a factored quadratic be set equal to zero before using the zero-product property?

The zero-product property applies specifically to products equal to zero. Therefore, all terms must first be moved to one side so the equation has the form $$(…)(…)=0.$$

How do you factor $x^2+bx+c$ when the leading coefficient is 1?

Find two numbers whose product is $c$ and whose sum is $b$. These numbers produce the factors $(x+m)(x+n)$, which can then be set individually equal to zero.

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

Because $3(-2)=-6$ and $3+(-2)=1$, factor the equation as $(x+3)(x-2)=0$. The solutions are $x=-3$ and $x=2$.

Difference of squares

A difference of two perfect squares factors as $A^2-B^2=(A-B)(A+B)$. For example, $x^2-9=(x-3)(x+3)$.

Solve $x^2-25=0$ using the difference-of-squares pattern.

Factor as $(x-5)(x+5)=0$. The solutions are $x=5$ and $x=-5$.

How should a quadratic with a nonunit leading coefficient be factored by grouping?

For $ax^2+bx+c$, calculate $ac$, find two numbers whose product is $ac$ and whose sum is $b$, split the middle term, and factor by grouping. Then apply the zero-product property.

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

Since $4(9)=36$ and $3+12=15$, rewrite the equation as $4x^2+3x+12x+9=0$. This factors to $(4x+3)(x+3)=0$, giving $x=-3/4$ and $x=-3$.

What should be done first when solving a polynomial equation such as $-3x^3-5x^2-2x=0$ by factoring?

Factor out the greatest common factor first. Here, $-x(3x^2+5x+2)=0$, after which the remaining quadratic can be factored and all factors can be set equal to zero.

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

Factor the expression as $-x(3x+2)(x+1)=0$. The solutions are $x=0$, $x=-2/3$, and $x=-1$.

Square root property

If $x^2=k$, then $x=±\sqrt{k}$, provided the square root is interpreted over the real or complex numbers as appropriate. The $±$ is necessary because both a positive and negative number can have the same square.

Why is the $±$ sign required when solving $x^2=8$?

Both $x=\sqrt{8}=2\sqrt{2}$ and $x=-\sqrt{8}=-2\sqrt{2}$ square to 8. Thus the solutions are $x=±2\sqrt{2}$.

When is the square root method usually the most efficient way to solve a quadratic?

It is especially efficient when the equation has no linear term, such as $ax^2+c=0$, or when it is already written in the form $A(x-h)^2=k$.

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

Divide by 3 to get $(x-4)^2=5$. Taking square roots gives $x-4=±\sqrt{5}$, so $x=4±\sqrt{5}$.

Completing the square

Completing the square transforms a quadratic into a perfect-square form, allowing the square root property to be used. The goal is to create an expression such as $(x+h)^2=k$.

What quantity must be added to $x^2+bx$ to complete the square?

Add $(b/2)^2$ to both sides of the equation. Then $x^2+bx+(b/2)^2$ factors as $(x+b/2)^2$.

What must be done before completing the square if the coefficient of $x^2$ is not 1?

Divide every term in the equation by the coefficient of $x^2$ so that the quadratic term has coefficient 1. Only then should $(b/2)^2$ be added to both sides.

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

Rearrange to $x^2+4x=-1$, add 4 to both sides, and obtain $(x+2)^2=3$. Therefore, $x=-2±\sqrt{3}$.

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

Rearrange to $x^2-3x=5$ and add $9/4$ to both sides: $(x-3/2)^2=29/4$. The solutions are $x=(3±\sqrt{29})/2$.

Quadratic formula

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

What procedure reduces substitution errors when using the quadratic formula?

First put the equation in standard form, identify $a$, $b$, and $c$, and substitute using parentheses around negative values. Then evaluate the discriminant and simplify both $±$ cases.

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

Here $a=1$, $b=5$, and $c=1$. Substitution gives $x=(-5±\sqrt{21})/2$.

How does a negative discriminant affect the solutions of a quadratic equation?

If $b^2-4ac<0$, the equation has two nonreal complex conjugate solutions. Using $\sqrt{-k}=i\sqrt{k}$, the solutions can be written with the imaginary unit $i$.

Solve $x^2+x+2=0$ over the complex numbers.

The discriminant is $1-8=-7$. Therefore, $x=(-1±\sqrt{-7})/2=(-1±i\sqrt{7})/2$.

Discriminant

The discriminant is $D=b^2-4ac$, the expression under the radical in the quadratic formula. It predicts the number and type of solutions without requiring the equation to be solved.

What does $D=0$ indicate for a quadratic equation?

There is one real rational solution counted twice, called a double root. Algebraically, the parabola touches the $x$-axis at one point.

What does $D>0$ indicate for a quadratic equation?

There are two distinct real solutions. If $D$ is a perfect square, both solutions are rational; if it is not, both are irrational.

What does $D<0$ indicate for a quadratic equation?

There are two nonreal complex conjugate solutions and no real solutions. Consequently, the corresponding parabola has no real $x$-intercepts.

Classify the solutions of $9x^2-30x+25=0$ using the discriminant.

The discriminant is $(-30)^2-4(9)(25)=0$. Thus there is one rational double solution.

How are the solutions of $ax^2+bx+c=0$ related to the graph of $y=ax^2+bx+c$?

The real solutions are the $x$-intercepts, or zeros, of the parabola. If the discriminant is negative, the graph has no real $x$-intercepts.

Why is only the positive square root used for a physical length?

The algebraic equation may produce both $±\sqrt{k}$, but a length represents a magnitude and cannot be negative. Therefore, the positive root is selected when interpreting the result physically.

How can a real-world condition be modeled as a quadratic equation?

Translate the situation into an equation, substitute the given target value, and rearrange into standard form $ax^2+bx+c=0$. Solve using factoring, square roots, completing the square, or the quadratic formula, then reject solutions that violate the context.

A rectangle has width $w$, length $w+10$, and area $119$. What quadratic equation models its dimensions?

The area equation is $w(w+10)=119$, which rearranges to $w^2+10w-119=0$. Solving gives $w=7$ or $w=-17$; only $w=7$ is physically valid, so the length is 17.

How can a quadratic model be used to find when a quantity reaches a specified value?

Set the quadratic expression equal to the specified value and rearrange to zero. Solve the resulting quadratic, then apply any domain restrictions such as nonnegative time or a stated interval.

Why might a quadratic model produce two mathematically valid times but only one valid answer in context?

A quadratic can intersect a target value twice, but the situation may restrict the variable to a particular interval or require a nonnegative time. Solutions outside that domain must be discarded.

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

By the relationships between roots and coefficients, the sum of the roots is $-b/a$. This is often called a Vieta relationship.

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

The product of the roots is $c/a$. Together with the sum $-b/a$, this can help check solutions or construct a quadratic from known roots.

What is a reliable procedure for solving an equation containing a radical?

Isolate the radical on one side, raise both sides to the power that removes the radical, solve the resulting equation, and check every candidate in the original equation.

Why must solutions to radical equations be checked in the original equation?

Squaring or applying another even power can introduce extraneous solutions. A candidate is valid only if it satisfies the original radical equation.

How can an equation with a rational exponent be rewritten to solve it?

Rewrite $x^{m/n}$ as the $n$th root of $x^m$, or raise both sides to a reciprocal power when permitted. Account for domain restrictions and verify solutions in the original equation, especially when an even root is involved.