Cracking the SAT subject test in math 2, 2nd edition

Cracking the SAT Subject Test in Math 2, 2nd Edition It’s possible to tell quickly, without going all the way through the quadratic formula, how many roots an equation has The part of the quadratic fo[.]

It’s possible to tell quickly, without going all the way through the quadratic formula, how many roots an equation has The part of the

quadratic formula under the radical, b2 − 4ac, is called the discriminant.

The value of the discriminant gives you the following information about a quadratic equation:

• If b2 − 4ac > 0, then the equation has two distinct real roots.

• If b2 − 4ac = 0, then the equation has one distinct real root and is a

perfect square Actually, it has two identical real roots, which ETS will call a “double root.”

• If b2 − 4ac < 0, then the equation has no real roots Both of its roots

are imaginary

DRILL 15: THE QUADRATIC FORMULA

In the following exercises, first use the discriminant to find how many roots each equation has and whether those roots are real or imaginary Next, use the quadratic formula to find the root or roots of each of the following To find imaginary roots, remember that (see Chapter 12

on imaginary numbers) The answers can be found in Part IV

1 x2 − 7x + 5 = 0

3 s2 − 6s + 4 = 0

4 x2 − 2 = 0

5 n2 + 5n + 6.25 = 0

6 x2 + 9 = 0

GRAPHING CALCULATOR TO THE RESCUE!

On the TI-84, there are tools you can use to solve certain tricky algebra questions For many problems with one variable, graphing the equation and using one of the following tools can be much easier (and safer!) than solving the problem algebraically

Finding Real Roots Using the Zero Function

You can use the zero function under the CALC menu to find the zeroes of

a function For example, let’s say you wanted to find the roots of

, when x ≠ 3 Input the function under the y = menu

(be careful using parentheses with the fraction) Next, press 2ND->TRACE

to access the CALC menu, and choose 2: zero The calculator will graph the function and lead you to this screen:

If you can’t see the graph clearly, use the WINDOW function and change the Xmax, Xmin, Ymax, and Ymin values Now, to select the smallest zero, use the arrows to select a point to the left of the smallest zero and hit ENTER This sets the “left bound.” Then move the cursor to a point to the right of the first zero but before the second zero and hit ENTER to select the “right bound.” Once you’ve selected the right bound you find yourself at this screen:

You find that f(x) = 0 at (−1.414, 0) Occasionally, you will not find an exact zero; sometimes the y value will be something vanishingly small This is due to how the TI series creates graphs (by calculating a series of y values based on a series of x values); however, for the purposes of the

SAT Subject Test in Math 2, “close enough” will get you down to one multiple-choice response

Finding Points of Intersection Using the

Intersect Function

You can use the intersect function under the CALC menu to find the points of intersection of two graphs This is useful when you want to find the solutions of two equations or functions which only have one variable each For example, let’s say you wanted to find the points of intersection

of 2x3 − 10x + 6 and + 2 First, put each function into the y = menu.

Next, use 2ND->ZOOM to access the FORMAT menu and make sure that CoordOn and ExprOn are enabled

Go to 2ND->TRACE to access the CALC menu, and go to 5: intersect Your calculator will graph both functions; if you cannot see the points of intersection, resize your window using the WINDOW function Once your calculator has finished graphing the functions, you will see the following screen:

You can use the up and down arrows to change which function you choose as the first function The function is listed at the top of the screen Once you’ve selected the correct function, press ENTER and you will be prompted to select the second function Once again, you can choose any

function that you have under the “y =” menu Press ENTER and you will

see the following:

Your cursor will follow the function you identified as the “second curve.” Trace that function until you are near the leftmost point of intersection, then press ENTER:

You’ve found the point of intersection to be approximately (-2.46, 0.77) This is one of the solutions to these equations To find the other

solutions, repeat the process starting from CALC-> intersect and select the other points of intersection in the last step