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# 2.1: Methods for Solving Quadratic Functions

Difficulty Level: At Grade Created by: CK-12
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Practice Methods for Solving Quadratic Functions
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Can you measure height with a stopwatch? How could that be possible?

This lesson is all about quadratic functions, like the one used in Physics and described later in the lesson, which allows a good approximation of the height of a tall object to be calculated with time!

### Watch This

This lesson focuses on the use of multiple methods for solving quadratic equations. Of the methods discussed, the "Factoring to Solve" and "Completing the Square" methods commonly require more practice and explanation.

You may choose to watch both videos below, or only the one referring to the solution method you are less comfortable with.

Embedded Video:

a. SOLVE BY FACTORING:

b. SOLVE BY COMPLETING THE SQUARE:

### Guidance

A quadratic function can be described as:

Formally: A function f\begin{align*}f\end{align*} defined by f(x)=ax2+bx+c\begin{align*}f(x)=ax^{2}+bx+c\end{align*}, where a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*} are real numbers and a0\begin{align*}a \ne 0\end{align*}.
Informally: The defining characteristic of a quadratic function is that it is a polynomial whose highest exponent is 2.

There are several ways to write quadratic functions:

standard form, the form of the quadratic function above: f(x)=ax2+bx+c\begin{align*}f(x)=ax^{2}+bx+c\end{align*}
vertex form, commonly used for quick sketching: f(x)=a(xh)2+k\begin{align*}f(x)=a(x-h)^{2}+k\end{align*}
factored form, excellent for finding x-intercepts: f(x)=a(xr1)(xr2)\begin{align*}f(x)=a(x-r_{1})(x-r_{2})\end{align*}

You can convert between forms of quadratic functions using algebra and you will see that there are uses for each of these forms when working with quadratic functions.

When graphing, the \begin{align*}y-\end{align*}intercept of a quadratic function in standard form is \begin{align*}(0, c)\end{align*} and it is found by substituting \begin{align*}0\end{align*} for \begin{align*}x\end{align*} in \begin{align*}f(x)=ax^{2}+bx+c\end{align*}.

#### Example A

Solve \begin{align*}x^{2}-5x+6=0\end{align*} using the "Factor to Solve" method.

Solution

The factor method is based on writing the quadratic equation in factored form. That is, as a product of two linear expressions. So our equation maybe solved by the following way:

\begin{align*}x^{2}-5x+6 & = 0\\ (x-3)(x-2) & = 0\end{align*}

Recall, that

\begin{align*}a \cdot b=0 \ \text{if and only if} \ a=0 \ \text{and} \ b=0\end{align*}

This tells us that

\begin{align*}x-3=0\end{align*}

or

\begin{align*}x-2=0\end{align*}

which give the roots (or zeros)

\begin{align*}x=3\end{align*}

and

\begin{align*}x=2\end{align*}

In other words, the solution set is {2, 3}.

#### Example B

Solve \begin{align*}x^{2}-5x+6=0\end{align*} by the "Completing the Square" method.

Solution

To solve the above equation by completing the square, first move the “\begin{align*}c\end{align*}” term to the other side of the equation,

\begin{align*}x^{2}-5x=-6\end{align*}

Next, make the left-hand side a “perfect square” by adding the appropriate number. To do so, take one-half of the coefficient of \begin{align*}x\end{align*} (the \begin{align*}b\end{align*} coefficient) and square it and then add the result to both sides of the equation:

\begin{align*}b & = -5 \\ \frac{b}{2} & = \frac{-5}{2}\\ \left ( \frac{b}{2} \right ) ^2 & = \frac{25}{4}\end{align*}

Adding to both sides of the equation,

\begin{align*}x^{2}-5x+\frac{25}{4} & = -6 + \frac{25}{4}\\ x^2 -5x + \frac{25}{4} & = \frac{1}{4}\\ \left ( x-\frac{5}{2} \right )^2 & = \frac{1}{4}\end{align*}

this last equation can be easily solved by taking the square root of both sides,

\begin{align*}\left(x-\frac{5}{2}\right)=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}\end{align*}

Hence

\begin{align*}x=\pm\frac{1}{2}+\frac{5}{2}\end{align*}

and the solutions are

\begin{align*}x=3\end{align*}

and

\begin{align*}x=2\end{align*}

which are identical to answers of the factor method.

#### Example C

Solve \begin{align*}x^{2}-5x+6=0\end{align*} using the Quadratic Formula.

Solution

The quadratic formula works for finding the x-intercepts in all quadratic equations, therefore it is highly encouraged that you memorize the formula. The other methods can be much faster though, so it is well worth understanding each of the different methods.

Recall, if \begin{align*}ax^{2}+bx+c=0\end{align*} where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} are real numbers and \begin{align*}a\ne0\end{align*}, then the roots of the equation can be determined by the quadratic formula.

\begin{align*}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.\end{align*}

Here the coefficients are \begin{align*}a=1, b=-5,\end{align*} and \begin{align*}c=6\end{align*}.

Substituting into the quadratic formula, we get:

\begin{align*}x & = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}\\ & = \frac{5 \pm \sqrt{25-24}}{2}\\ & = \frac{5 \pm 1}{2}\\ & = 3 \ \text{or} \ 2\end{align*}

which is, again, identical to our two solutions above.

Were you able to solve the problem at the beginning of the lesson?

What formula could be used to calculate the height of an object based on the time required for an object to fall from the top?

One formula is \begin{align*}h = 1/2 gt^{2}\end{align*}

This is not completely accurate, as it ignores the effects of air resistance of an object, but offers a very close approximate for most common objects.

The value g is gravity. In Metric units this is \begin{align*}9.81 m/s/s\end{align*}

The value t is the time taken for the object to fall in seconds

The result h is the distance or the height in meters

### Vocabulary

Completing the Square: A common method for solving quadratic equations accomplished by making a perfect square trinomial on one side of the equation.

Factor to Solve: A common method for solving quadratic equations accomplished by factoring a trinomial into two binomials and identifying the values of x that make each binomial equal to zero.

Quadratic Formula The most reliable method of solving for x intercepts in a quadratic equation, though not always the most efficient, involves substituting the coefficients of each of the terms of the quadratic function into the formula: \begin{align*}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.\end{align*}

### Guided Practice

Questions

1) Solve each equation by factoring.

a. \begin{align*}x^{2}=64\end{align*}
b. \begin{align*}4x^{2}+7x=2\end{align*}
c. \begin{align*}4x^{2}-17x=-4\end{align*}

2) Solve each equation by completing the square

a. \begin{align*}x^{2}-4x+1=0\end{align*}
b. \begin{align*}\frac{2}{3}x^{2}-x+\frac{1}{3}=0\end{align*}
c. \begin{align*}x^{2}+2.8=4.7x\end{align*}

3) Solve each equation by using the quadratic formula

a. \begin{align*}0.4x^{2}+x-0.3=0\end{align*}
b. \begin{align*}25x^{2}+80x+61=0\end{align*}
c. \begin{align*}(z+6)^{2}+2z=0\end{align*}

1) Factoring to Solve:

a. \begin{align*}x = \sqrt{64} \therefore x = \pm8\end{align*}
b. \begin{align*}4x^{2} + 7x - 2 = 0\end{align*}
\begin{align*}0 = (4x - 1)(x + 2) \therefore x = \frac{1}{4}, -2\end{align*}
c. \begin{align*}4x^{2} - 17x + 4 = 0\end{align*}
\begin{align*}0 = (4x - 1)(x - 4) \therefore x = \frac{1}{4}, 4\end{align*}

2) Completing the Square:

a. \begin{align*}x^{2}-4x+4=3\end{align*}
\begin{align*}(x-2)(x-2)=3\end{align*}
\begin{align*}x = 2\pm\sqrt{3}\end{align*}
b. \begin{align*}x^{2}-\frac{3}{2}x+\frac{1}{2}=0\end{align*}
\begin{align*}x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}-\frac{1}{2}\end{align*}
\begin{align*}(x-\frac{3}{4})(x-\frac{3}{4}) = \frac{1}{16}\end{align*}
\begin{align*}x=\frac{1}{2}, 1\end{align*}
c. \begin{align*}x^{2} - 4.7x + 2.8 = 0\end{align*}
\begin{align*}x^2 - 4.7x + 2.35^{2} = 2.35^{2}-2.8\end{align*}
\begin{align*}(x-2.35)(x-2.35)=2.7225\end{align*}
\begin{align*}x-2.35=\pm1.65\end{align*}
\begin{align*}x=\frac{7}{10}, 4\end{align*}

a. \begin{align*}x = \frac{-1\pm\sqrt{1^{2}-4 \cdot 0.4 \cdot (-0.3)}}{2\cdot0.4}\end{align*}
\begin{align*}x=0.275, -2.78\end{align*}
b. \begin{align*}x = \frac{-80\pm\sqrt{80^{2}-4\cdot25\cdot61}}{2\cdot25}\end{align*}
\begin{align*}x = -\frac{8}{5}\pm\frac{\sqrt{3}}{5}\end{align*}
c. \begin{align*}z^{2}+14z+36=0 \therefore x = \frac{-14\pm\sqrt{14^{2}-4\cdot1\cdot36}}{2\cdot1}\end{align*}
\begin{align*}x = -7\pm\sqrt{13}\end{align*}

### Practice

1. List three ways to write a quadratic function. a) b) c)

1. \begin{align*}m^2 - 5m - 14 = 0\end{align*}
2. \begin{align*}b^2 - 4b + 4 = 0\end{align*}
3. \begin{align*}4b^2 + 8b + 7 = 0\end{align*}
4. \begin{align*}2m^2 - 7m - 13 = -10\end{align*}
5. \begin{align*}5r^2 = 80\end{align*}
6. \begin{align*}k^2 - 31 - 2k = -6 - 3k^2 - 2k\end{align*}
1. If you have an equation with a power of 4, explain how you could solve it using the quadratic formula.

Solve by completing the square:

1. \begin{align*}2x^2 - 12x + 26 = 10\end{align*}
2. \begin{align*}x^2 - 12x + 29 = -3\end{align*}
3. \begin{align*}7x^2 - 14x -64 = -8\end{align*}

Solve by the most efficient method:

1. \begin{align*}x^4 + 13x^2 + 36 = 0\end{align*}
2. \begin{align*}x^4 + 16x^2 -225 = 0\end{align*}
3. \begin{align*}\frac{1}{4}x^2 - \frac{1}{3}x + 1 = 0\end{align*}
4. \begin{align*}\frac{2}{7}c^2 - \frac{1}{2}c - \frac{3}{14} = 0\end{align*}

In the quadratic formula \begin{align*} b^2 - 4ac\end{align*} is called the discriminant. The values of the discriminant tell us the nature of the solutions or roots of a quadratic equation, \begin{align*}ax^2 + bx + c = 0\end{align*}

1. What value(s) of the discriminant result in two unique real solutions?
2. What value(s) of the discriminant result in one unique real solutions?
3. What value(s) of the discriminant result in two uniqueimaginary solutions?

### Vocabulary Language: English

Completing the Square

Completing the square is a common method for rewriting quadratics. It refers to making a perfect square trinomial by adding the square of 1/2 of the coefficient of the $x$ term.

Factor to Solve

"Factor to Solve" is a common method for solving quadratic equations accomplished by factoring a trinomial into two binomials and identifying the values of $x$ that make each binomial equal to zero.

factored form

The factored form of a quadratic function $f(x)$ is $f(x)=a(x-r_{1})(x-r_{2})$, where $r_{1}$ and $r_{2}$ are the roots of the function.

Factoring

Factoring is the process of dividing a number or expression into a product of smaller numbers or expressions.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

A quadratic function is a function that can be written in the form $f(x)=ax^2 + bx + c$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.

Roots

The roots of a function are the values of x that make y equal to zero.

standard form

The standard form of a quadratic function is $f(x)=ax^{2}+bx+c$.

Vertex form

The vertex form of a quadratic function is $y=a(x-h)^2+k$, where $(h, k)$ is the vertex of the parabola.

Zeroes of a Polynomial

The zeroes of a polynomial $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.

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Date Created:
Nov 01, 2012