Math: Quadratics and Other Curves

This is Unit Eleven of ACT Preparation: quadratics and other curves. Unit ten defined the function; this unit studies the curves that functions make, starting with the parabola. One habit runs through it: let the form of an expression choose the method, then check the answer against the original. By the end of this part you will be able to solve a quadratic four ways. To count its real solutions with the discriminant. To use the sum and product of its roots. To solve radical equations and discard extraneous answers. And to find a parabola's vertex and solve a maximum problem.

Eight entries on solving quadratic equations, and choosing among the methods.

If a product is zero.

ACT 1001 The Zero-Product Property

If a product is zero, a factor is zero: set the quadratic equal to zero, factor, and set each factor to zero. X squared minus 5x minus fourteen factors as x minus seven, times x plus two, so x is seven or negative two. Get zero alone on one side first.

Taking a square root.

ACT 1002 The Square Root Property

When a squared expression stands alone, take the square root of both sides, keeping both signs. X minus three, squared, equals twenty-five gives x minus three equals plus or minus five, so x is eight or negative two. Forgetting the minus sign loses a solution.

Completing the square.

ACT 1003 Completing the Square

To complete the square in x squared plus b x, add half of b, squared, to both sides. X squared plus 6x equals seven becomes x plus three, squared, equals sixteen, so x is one or negative seven. The leading coefficient must be one first; divide through if it is not.

The formula that always works.

ACT 1004 The Quadratic Formula

The quadratic formula gives the solutions of a x squared plus b x plus c equals zero: negative b, plus or minus the square root of b squared minus 4ac, all over 2a. For 2x squared plus 3x minus two, that is negative three plus or minus five, over four: one half or negative two. Divide the whole numerator by 2a.

Counting solutions before finding them.

ACT 1005 The Discriminant

The discriminant, b squared minus 4ac, counts the real solutions: positive gives two, zero gives one repeated solution, negative gives none real. For x squared plus 4x plus five, sixteen minus twenty is negative four, so the parabola misses the x-axis.

Choosing a method, step by step.

ACT 1006 Choosing a Method for a Quadratic

If a squared expression stands alone or there is no x term, use square roots; otherwise set the equation to zero and try to factor; if factoring stalls, use the formula, which never fails. 4x squared equals forty-nine gives plus or minus seven halves. X squared plus 2x minus fifteen factors, giving negative five or three. X squared plus 2x minus four does not, so the formula gives negative one plus or minus root five. Every method gives the same solutions; the choice only saves time.

Using the formula, step by step.

ACT 1007 Solving With the Quadratic Formula

Write the equation in standard form and read a, b and c with their signs, compute the discriminant, substitute, then simplify the radical and reduce. 3x squared equals 2x plus four becomes a equals three, b negative two, c negative four; the discriminant is fifty-two; and the solutions reduce to one plus or minus root thirteen, over three. When b is negative, negative b is positive, so the formula starts with plus two.

Using the discriminant, step by step.

ACT 1008 Using the Discriminant

Write the quadratic in standard form, write b squared minus 4ac, and set it greater than, equal to, or less than zero as the question requires. For exactly one real solution of x squared plus k x plus nine, k squared minus thirty-six equals zero, so k is six or negative six. Exactly one solution makes the equation a perfect square, a fast check: x plus three, squared.

Four entries on the roots' sum and product, and on radical equations.

What the roots add and multiply to.

ACT 1009 Sum and Product of the Roots

For a x squared plus b x plus c equals zero, the roots sum to negative b over a and multiply to c over a. 2x squared minus 10x plus twelve has roots summing to five and multiplying to six: two and three. Mind the sign of the sum.

Radical equations.

ACT 1010 Radical Equations and Extraneous Solutions

To solve a radical equation, isolate the radical, raise both sides to the root's index, solve, and check every answer in the original. The square root of x plus six equals x gives three or negative two, and only three checks. Squaring can create extraneous solutions: the square root of four is two, never negative two.

Rational exponents in equations.

ACT 1011 Equations With Rational Exponents

To solve x to the m over n equals k, raise both sides to the reciprocal power and check. X to the three halves equals twenty-seven gives x equals twenty-seven to the two thirds, nine. When m is even, as in x to the two thirds equals four, include both signs: plus or minus eight.

A radical equation, step by step.

ACT 1012 Solving a Radical Equation

Isolate the radical, raise both sides to the index, solve, and substitute every answer into the original. The square root of 2x plus seven, plus four, equals x becomes the root equals x minus four; squaring gives x squared minus 10x plus nine equals zero, so nine or one. Nine checks; one gives seven, not one, and is discarded. Square the whole side: x minus four, squared, is x squared minus 8x plus sixteen.

Six entries on the parabola: its vertex, its axis, and its extreme value.

The graph of a quadratic.

ACT 1013 The Parabola

The graph of a quadratic is a parabola, opening up when a is positive and down when a is negative. Its turning point is the vertex, and its y-intercept is c. Negative x squared plus four opens down, with vertex zero, four and zeros at two and negative two.

The axis of symmetry.

ACT 1014 Axis of Symmetry

A parabola is symmetric about the vertical line through its vertex, x equals negative b over 2a. For x squared minus 6x plus five, the axis is x equals three and the vertex three, negative four. The zeros sit equally far on either side: one and five.

Vertex form.

ACT 1015 Vertex Form

In vertex form, a times x minus h, squared, plus k, the vertex is h, k. Two times x plus one, squared, minus eight has vertex negative one, negative eight and opens up. The sign inside flips: x plus one means h is negative one.

The highest or lowest value.

ACT 1016 Maximum and Minimum

A parabola's vertex gives its extreme value: a minimum when it opens up, a maximum when it opens down. Negative 2x squared plus 8x has its maximum at x equals two, where the value is eight. Answer the question asked: where the maximum occurs, or what it is.

Reaching vertex form, step by step.

ACT 1017 Completing the Square to Vertex Form

Factor a out of the x terms, add and subtract the square of half the x coefficient inside, move the subtracted amount outside multiplied by a, then write the square and combine constants. 2x squared minus 12x plus seven becomes two times x minus three, squared, minus eleven, with vertex three, negative eleven. The amount moved outside is multiplied by a: subtracting nine instead of eighteen gives a wrong vertex at three, negative two.

A maximum in a word problem, step by step.

ACT 1018 Finding a Maximum in a Word Problem

Write the quantity to optimize as a function, put it in standard form and confirm the direction, find the vertex, and answer with units. With 120 meters of fencing on three sides of a pen against a barn, the area is negative 2x squared plus 120x; its vertex is a width of thirty meters, a length of sixty, and a greatest area of 1,800 square meters. Check the answer fits: the width must leave a positive length.

Look at what you can now do. You can solve a quadratic by the method its form invites, with the formula always available. You can count real solutions with the discriminant, even with an unknown constant. You can use the roots' sum, negative b over a, and product, c over a. You can solve radical equations and discard the extraneous answers by checking each one. And you can find a parabola's vertex, rewrite it in vertex form, and solve a maximum problem. Part two turns to polynomials, rational functions and variation.