System of Nonlinear Equations

June 11, 2019/in Algebra, CLEP, College Algebra, graphs of circles, graphs of linear equations, graphs of systems of nonlinear equations, Math Test Prep, system of equations, systems of nonlinear equations/by Jill Hacker

System of Nonlinear Equations

Question

Find the value or values of x that are solutions to this system of equations:

$x^2+y^2=5$ $x+y=1$

Solution

If the equations look familiar enough so that you can sketch them by hand, then doing so may help you get your arms around the problem. (The calculator provided for the exam does not have graphing capability, so if you want to see a graph you pretty much need to make it yourself. Learn more about it here.)

A very rough sketch will do. I think my sketch (left) is pretty crude. For this problem, the first equation is for a circle with center at the origin and radius equal to the square root of 5. The square root of 5 is about 2.2, so the circle goes through points (2.2, 0); (0, 2); (-2.2, 0); and (0, -2.2), approximately.

The second is for a line with a slope of -1 and intercepts at $x=1$ and $y=1$ . The solutions to the system lie at the points of intersection of the graphs, one in the second quadrant (where x is negative and y is positive) and the other in the fourth quadrant (where x is positive and y is negative).

The sketch is just to help you get your arms around the problem. If it doesn’t work for you, you can skip it — it’s not essential.

You can solve using substitution. If $x+y=1$ then $y=1-x$ ; substitute $1-x$ for $y$ in the equation $x^2+y^2=5.$

$x^2+(1-x)^2=5$
$2x^2-2x-4=0$	Simplify: Multiply out, subtract 5 from each side, and collect like terms.
$x^2-x-2=0$	Divide both sides by 2.
$(x-2)(x+1)=0$	Factor
$x=-1, 2$	Set each factor equal to 0 and solve.

That’s all you need to answer the question.

To check your answers, use the equation $y=1-x$ to find the y-values that correspond to the x-values you found, and then substitute x- and y-values into the equation $x^2+y^2=5.$

This question is about nonlinear systems of equations. (The whole system gets labeled nonlinear even though one of the equations is linear.)

What’s up (more), the parabola or the line?

April 2, 2019/in Algebra, CLEP, College Algebra, finding x-intercepts, finding zeros, functions, graphing, linear equations, Math Test Prep, parabolas and lines/by Jill Hacker

What’s up (more), the parabola or the line?

Question

The graph to the left shows the function f(x), a parabola – that is, a quadratic function. Not shown on the graph is function $g(x) = cx,$ in which c is a negative constant.

One statement below is true. Which one?

$f(0)<g(0)$
$f(a)<g(0)$
$f(a)<g(a)$
$f(a)<g(b)$
$f(b)<g(b)$

Solution

What’s going on? $g(x) = cx$ is the equation of a straight line. The line will cross the parabola in a couple of places and those intersections will determine which function is higher in various places. But all you know about function $g$ is that $g(x) = cx$ and that c is negative.

Think of a line’s basic equation: $y=mx+b,$ where b is the line’s y-intercept – that is, y’s value where $x=0$ . In this case, because $y=cx +$ nothing, the y-intercept is 0. That means the line goes through the origin.

Also, you’re told that c is negative. c is in the role of m in the equation $y=mx+b;$ that is, it’s the slope. The negative slope means this line goes down as it goes to the right.

How steep is the line? You are given no information about that. Take a guess.

The brown line in the graph to the right shows what function g may look like. It may be steeper or less steep than shown, but it will cross the x-axis at the origin, and it will go down as it goes to the right, as shown.

Let’s go through all the possible answers to find the one true statement.

$f(0)<g(0)$ . At $x=0$ , as you can see from the graph, the parabola is above the x-axis, which means f(0) is a positive value. The line goes right through the origin – so $g(0)=0$ . Thus $f(0) > g(0),$ which means it cannot be true that $f(0)<g(0)$ .
$f(a)<g(0)$ . At $x=a$ , function f crosses the x-axis. So $f(a)=0$ . At $x=0$ , function g crosses the x-axis. So $g(0)=0$ . That means f(a) is equal to g(0), so it’s not possible that f(a) is less than g(0).
$f(a)<g(a)$ . What’s happening at $x=a$ ? Function f crosses the x-axis there, so $f(a)=0$ ; and function g has a positive value. That means g is more than f, so it must be that $f(a)<g(a)$ . Looks like that’s the answer, but let’s still check the others.
$f(a)<g(b)$ . We have established that $f(a)=0$ . And since g(x) goes through the origin and slopes down, g must be less than zero at any x-value that is greater than 0 – like, for example, b. That means g is less than f, so it is not possible that f is less than g.
$f(b)<g(b)$ . As you can see from the graph, $f(b)=0$ . And, as we said in response to answer #4, g(b) is less than 0. So f(b) must be more than g(b) – and it’s not possible that $f(b)<g(b)$ .

So answer 3, and only answer 3, works.

I found this question intimidating, because it seems so difficult to know anything about function g. But once you look closely, it’s not so bad. The key is to notice that g takes on positive values to the left of $x=0$ and negative values to the right of $x=0$ .

Rational Roots Test

November 13, 2018/in CLEP, factor theorem, finding roots, finding x-intercepts, finding zeros, Math Test Prep, polynomials, rational roots test, rational zeros test, roots, synthetic division, zeros/by Jill Hacker

Factoring a Sum of Cubes

October 30, 2018/in CLEP, factoring, factoring a difference of cubes, factoring a sum of cubes, Math Test Prep, volume, word problems/by Jill Hacker

Factoring a Sum of Cubes

Problem

There is a box with a volume of $x^3-64$ cubic inches, a rectangular base, and a height of $x-4$ inches. What is the area of the box’s base? Use the formula $V=bh,$ where $V$ is volume, $B$ is area of the base, and $h$ is height. Give your answer as an expression in simplest form and in terms of $x.$

Solution

You’re told that the box’s volume is equal to the area of its base times its height: $V=Bh.$ Then the area of this box’s base is equal to its volume divided by its height: $B=V/h.$

$B=\frac{V}{h}=\frac{x^3-64}{x-4}$

To get your answer into simplest form, factor the sum of cubes in the numerator and then carry out any possible cancellation between numerator and denominator. To do that factor the numerator. The numerator is the difference of two cubes, which factors like this:

$a^3-b^3=\left (a-b \right )\left (a^2+ab+b^2 \right )$

Apply that to this case:

$a=x; b=4$

Then

$x^3-4^3=\left ( x-4\right )\left ( x^2+4x+16\right )$

And the area of the base is

$\frac{\left (x-4 \right )\left ( x^2+4x+4 \right )}{x-4}=\bold{x^2+4x+16}$

The hard part of this problem is having the formula for the difference of cubes memorized. Alternatively, you can do polynomial long division or synthetic division. Synthetic division for this problem, $\frac{x^3-64}{x-4},$ should look like this:

Same result.

You may be asked about the sum, rather than the difference, of cubes. That formula is:

$a^3+b^3=\left ( a+b\right )\left ( a^2-ab+b^2\right )$