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Polynomial

FactorsQuadratic graphs

Completing the square

Here you will learn about the quadratic formula and how you can use it to solve quadratic equations.

Students will first learn about the quadratic formula as part of algebra in high school.

The **quadratic formula** is used to solve quadratic equations by finding the roots, x.

The quadratic formula is: x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a}

By using the general form of a **quadratic equation**, a x^{2}+b x+c=0, you can substitute the values of a, b and c into the quadratic formula to calculate x (the solution(s) for the quadratic formula).

The \pm in front of the square root means ‘plus or minus.’ This accounts for the positive root and negative root.

Quadratic equations normally have two solutions, so you need to use the formula twice, once with a + and once with a –.

x=\cfrac{-b+\sqrt{b^2-4ac}}{2a}\quad x=\cfrac{-b-\sqrt{b^2-4ac}}{2a}

To find out how many real roots a quadratic equation has, use the expression under the square root in the quadratic formula → b^2-4 a c. This is called the **discriminant**.

- If b^{2}-4ac > 0, this means there are two solutions (these are real numbers).

- If b^{2}-4ac=0 , this means there is one solution (which is a real number).

- If b^{2}-4ac < 0, this means there are no real solutions, here the solutions will be imaginary numbers.

For example,

x^2+3x+2=0Let’s see how many real solutions this quadratic equation has.

\begin{aligned} & b^2-4 ac \\\\ & =3^2-4 \times 1 \times 2 \\\\ & =9-8 \\\\ & =1 \end{aligned}Because the discriminant is greater than 0, there are two solutions (real numbers).

Now, let’s solve for the two solutions.

\begin{aligned} & x=\cfrac{-3+\sqrt{3^2-4 \times 1 \times 2}}{2 \times 1} \\\\ & x=\cfrac{-3+\sqrt{1}}{2 \times 1} \\\\ & x=\cfrac{-3+1}{2} \\\\ & x=\cfrac{-2}{2} \\\\ & x=-1 \end{aligned} \quad \begin{aligned} & x=\cfrac{-3-\sqrt{3^2-4 \times 1 \times 2}}{2 \times 1} \\\\ & x=\cfrac{-3-\sqrt{1}}{2 \times 1} \\\\ & x=\cfrac{-3-1}{2} \\\\ & x=\cfrac{-4}{2} \\\\ & x=-2 \end{aligned}

You can check by graphing the solution. Remember, the solutions are the x -intercepts.

How does this relate to high school math?

**Algebra – Reasoning with Equations and Inequalities (HSA.REI.B.4b)**

Solve quadratic equations in one variable. Solve quadratic equations by inspection ( e.g., for x^2=49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation.

Recognize when the quadratic formula gives complex solutions and write them as a \pm bi for real numbers a and b.

Use this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEIn order to use the quadratic formula:

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.****Substitute these values into the quadratic formula.****Solve the equation with a \bf{+}, and then with a \bf{-}.**

Solve x^{2}-8x+15=0.

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.**

2**Substitute these values into the quadratic formula.**

Using parentheses will help to make the calculation clear.

3**Solve the equation with a \bf{+}, and then with a** ** \bf{-}. **

\begin{aligned} & x=\cfrac{-(-8)+\sqrt{(-8)^2-4(1)(15)}}{2(1)} \\\\ & x=\cfrac{8+\sqrt{64-60}}{2} \\\\ & x=\cfrac{8+2}{2} \\\\ & x=5 \end{aligned}\quad \begin{aligned} & x=\cfrac{-(-8)-\sqrt{(-8)^2-4(1)(15)}}{2(1)} \\\\ & x=\cfrac{8-\sqrt{64-60}}{2} \\\\ & x=\cfrac{8-2}{2} \\\\ & x=3\end{aligned}

When you plot the graph of the quadratic equation you get a special ‘U’ shaped curve called a parabola.

By using graphing techniques such as this, you can see that the roots of the quadratic formula are where the quadratic graph **crosses the \textbf{x} -axis** or the x -intercepts.

You also can check that the solutions are correct by substituting them back into the quadratic equation.

Solve 2x^{2}+5x+3=0.

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.**

\begin{aligned}
& a=2 \\\\
& b=5 \\\\
& c=3
\end{aligned}

**Substitute these values into the quadratic formula.**

\begin{aligned}
& x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\\\
& x=\frac{-(5)\pm\sqrt{(5)^2-4(2)(3)}}{2(2)}
\end{aligned}

**Solve the equation with a** ** \bf {+}, and then with a \bf {-}. **

\begin{aligned} & x=\cfrac{-(5)+\sqrt{(5)^2-4(2)(3)}}{2(2)} \\\\ & x=\cfrac{-5+\sqrt{25-24}}{4} \\\\ & x=\cfrac{-5+1}{4} \\\\ & x=-1 \end{aligned}\quad \begin{aligned} & x=\cfrac{-(5)-\sqrt{(5)^2-4(2)(3)}}{2(2)} \\\\ & x=\cfrac{-5-\sqrt{25-24}}{4} \\\\ & x=\cfrac{-5-1}{4} \\\\ & x=-1.5\end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x -axis.

You can also check that the solutions are correct by substituting them back into the quadratic equation.

Solve -6x^2+x+2=0.

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.**

\begin{aligned}
& a=-6 \\\\
& b=1 \\\\
& c=2
\end{aligned}

**Substitute these values into the quadratic formula.**

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(1) \pm \sqrt{(1)^2-4(-6)(2)}}{2(-6)}
\end{aligned}

**Solve the equation with a \bf {+}, and then with a** ** \bf {-}. **

\begin{aligned} & x=\cfrac{-(1)+\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1+\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1+7}{-12} \\\\ & x=-\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-(1)-\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1-\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1-7}{-12} \\\\ & x=\cfrac{2}{3}\end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x -axis.

You can also check that the solutions are correct by substituting them back into the quadratic equation.

Solve x^2+14=9 x.

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.**

In order to solve the following equation, x^2+14=9x, all terms of the equation need to be on the left-hand side of the equation.

\begin{aligned}
& x^2+14=9 x \\\\
& -9 x-9 x \\\\
& x^2-9 x+14=0
\end{aligned}

\begin{aligned}
& a=1 \\\\
& b=-9 \\\\
& c=14
\end{aligned}

**Substitute these values into the quadratic formula.**

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-9) \pm \sqrt{(-9)^2-4(1)(14)}}{2(1)}
\end{aligned}

**Solve the equation with a \bf {+}, and then with a \bf {-}. **

\begin{aligned} & x=\cfrac{-\left(-9)+\sqrt{(-9)^2-4(1)(14)}\right.}{2(1)} \\\\ & x=\cfrac{9+\sqrt{81-56}}{2} \\\\ & x=\cfrac{9+5}{2} \\\\ & x=7 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-9)-\sqrt{(-9)^2-4(1)(14)}}{2(1)} \\\\ & x=\cfrac{9-\sqrt{81-56}}{2} \\\\ & x=\cfrac{9-5}{2} \\\\ & x=2\end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x -axis.

You can also check that the solutions are correct by substituting them back into the quadratic equation.

Solve x^2-x+1=0.

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.**

\begin{aligned}
& a=1 \\\\
& b=-1 \\\\
& c=1
\end{aligned}

**Substitute these values into the quadratic formula.**

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-1) \pm \sqrt{(-1)^2-4(1)(1)}}{2(1)}
\end{aligned}

**Solve the equation with a \bf {+}, and then with a \bf {-}. **

\begin{aligned} & x=\cfrac{-(-1)+\sqrt{(-1)^2-4(1)(1)}}{2(1)} \\\\ & x=\cfrac{1+\sqrt{1-4}}{2} \\\\ & x=\cfrac{1+\sqrt{3} i}{2} \\\\ & x=0.5+0.866 i \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-1)-\sqrt{(-1)^2-4(1)(1)}}{2(1)} \\\\ & x=\cfrac{1-\sqrt{1-4}}{2} \\\\ & x=\cfrac{1-\sqrt{3} i}{2} \\\\ & x=0.5-0.866 i \end{aligned}

Note, since \sqrt{1-4}=\sqrt{-3}, you use a complex number to represent the square root.

The roots are imaginary numbers, because this parabola does not cross the x -axis.

Solve 2x^2-7 x+19=0.

**Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.**

\begin{aligned}
& a=2 \\\\
& b=-7 \\\\
& c=19
\end{aligned}

**Substitute these values into the quadratic formula.**

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-7) \pm \sqrt{(-7)^2-4(2)(19)}}{2(2)}
\end{aligned}

**Solve the equation with a \bf{+}, and then with a \bf{-}. **

\begin{aligned} & x=\cfrac{-(-7)+\sqrt{(-7)^2-4(2)(19)}}{2(2)} \\\\ & x=\cfrac{7+\sqrt{49-152}}{4} \\\\\ & x=\cfrac{7+\sqrt{103 i}}{4} \\\\ & x=1.75+2.537 i \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-7)-\sqrt{(-7)^2-4(2)(19)}}{2(2)} \\\\ & x=\cfrac{7-\sqrt{49-152}}{4} \\\\ & x=\cfrac{7-\sqrt{103 i}}{4} \\\\ & x=1.75-2.537 i \end{aligned}

Note, since \sqrt{49-152}=\sqrt{-103}, you use a complex number to represent the square root.

The roots are imaginary numbers, because this parabola does not cross the x -axis.

- Let students explore other strategies for finding roots before introducing the quadratic formula, such as graphing or factoring the quadratic equation.

- Once students start using the quadratic formula, have them explain the meaning of their answers. This ensures that they do not forget what they are solving for. Better yet, be sure to incorporate many real world examples of the quadratic equation and require students to explain the roots within the context of the problem.

**Forgetting to complete both calculations of the \textbf{x} value**When writing the quadratic formula, always write it with the plus and minus symbol. The quadratic formula needs to be solved twice, once with the minus sign and once with the plus sign. This is because there is a positive and negative solution for the square root of a number.

**Not including the sign in front of \textbf{a, b} and \textbf{c}**When writing the coefficients of the variables for the values of a, b and c, you need to include the sign in front but not the variable.

For example,x^{2}-8x+15=0

so here b=-8, NOT 8 or -8x

**Typing incorrectly into a calculator**

Mistakes can be made when typing into a calculator.

For example, for2x^{2}+5x+3=0,

Use the fraction button and brackets to carefully type into the calculator

a=2,\qquad b=5,\qquad c=3

Use the arrow buttons to change the + on the top left of the calculation to a - to calculate the second solution.

**Forgetting that sometimes there are no real roots to a quadratic equation or quadratic function**For example,x^{2}-4 x+5=0

You can see that there are no values of x that give a y value of 0.

The graph does not cross the x -axis. The use of the quadratic formula will result in roots that are not real.

- Solving quadratic equation
- Factoring quadratic equations

1) Solve {x}^2+4x+3=0 using the quadratic formula.

x=3 \quad x=-1

x=-3 \quad x=1

x=-3 \quad x=-1

x=3 \quad x=1

Identify a, b and c.

a=1, \quad b=4, \quad c=3

Substitute these values into the quadratic formula.

\begin{aligned} & x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-4 \pm \sqrt{(4)^2-4(1)(3)}}{2(1)} \end{aligned}

Solve the equation with a +, and then with a -.

\begin{aligned} & x=\cfrac{-4+\sqrt{(4)^2-4(1)(3)}}{2(1)} \\\\ & x=\cfrac{-4+\sqrt{16-12}}{2} \\\\ & x=\cfrac{-4+2}{2} \\\\ & x=-1 \end{aligned} \quad \begin{aligned} & x=\cfrac{-4-\sqrt{(4)^2-4(1)(3)}}{2(1)} \\\\ & x=\cfrac{-4-\sqrt{16-12}}{2} \\\\ & x=\cfrac{-4-2}{2} \\\\ & x=-3 \end{aligned}

2) Solve {x}^2+3x-18=0 using the quadratic formula.

x=-6 \quad x=-3

x=-6 \quad x=3

x=6 \quad x=3

x=6 \quad x=-3

Identify a, b and c.

a=1, \quad b=3, \quad c=-18

Substitute these values into the quadratic formula.

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-3 \pm \sqrt{(3)^2-4(1)(-18)}}{2(1)}
\end{aligned}

Solve the equation with a +, and then with a -.

\begin{aligned}
& x=\cfrac{-3+\sqrt{(3)^2-4(1)(-18)}}{2(1)} \\\\
& x=\cfrac{-3+\sqrt{9-(-72)}}{2} \\\\
& x=\cfrac{-3+9}{2} \\\\
& x=3 \end{aligned} \quad \begin{aligned}
& x=\cfrac{-3-\sqrt{(3)^2-4(1)(-18)}}{2(1)} \\\\
& x=\cfrac{-3-\sqrt{9-(-72)}}{2} \\\\
& x=\cfrac{-3-9}{2} \\\\
& x=-6
\end{aligned}

3) Solve 2{x}^2-3x-2=0 using the quadratic formula.

x=-0.5 \quad x=2

x=-0.5 \quad x=-2

x=0.5 \quad x=2

x=0.5 \quad x=-2

Identify a, b and c.

a=2, \quad b=-3, \quad c=-2

Substitute these values into the quadratic formula.

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-2)}}{2(2)}
\end{aligned}

Solve the equation with a +, and then with a -.

\begin{aligned}
& x=\cfrac{-(-3)+\sqrt{(-3)^2-4(2)(-2)}}{2(2)} \\\\
& x=\cfrac{3+\sqrt{9-(-16)}}{4} \\\\
& x=\cfrac{3+5}{4} \\\\
& x=2
\end{aligned} \quad \begin{aligned}
& x=\cfrac{-(-3)-\sqrt{(-3)^2-4(2)(-2)}}{2(2)} \\\\
& x=\cfrac{3-\sqrt{9-(-16)}}{4} \\\\
& x=\cfrac{3-5}{4} \\\\
& x=-0.5
\end{aligned}

4) Solve -x^2+2 x+1=-11 using the quadratic formula.

x=-0.414, \quad x=2.414

x=-2.6055, \quad x=4.6055

x=-2.317, \quad x=4.317

x=1+3i \quad x=1-3i

First, convert the equation to the general form ax^2+b x+c=0.

\begin{aligned}
-x^2+2 x+1= & -11 \\\\
+11 \hspace{0.2cm} & +11 \\\\
-x^2+2 x+12= & 0
\end{aligned}

Identify a, b and c.

a=-1, \quad b=2, \quad c=12

Substitute these values into the quadratic formula.

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-2 \pm \sqrt{(2)^2-4(-1)(12)}}{2(-1)}
\end{aligned}

Solve the equation with a +, and then with a -.

\begin{aligned}
& x=\cfrac{-2+\sqrt{(2)^2-4(-1)(12)}}{2(-1)} \\\\
& x=\cfrac{-2+\sqrt{4-(-48)}}{-2} \\\\
& x=\cfrac{-2+7.211}{-2} \\\\
& x=-2.6055
\end{aligned} \quad \begin{aligned}
& x=\cfrac{-2-\sqrt{(2)^2-4(-1)(12)}}{2(-1)} \\\\
& x=\cfrac{-2-\sqrt{4-(-48)}}{-2} \\\\
& x=\cfrac{-2-7.211}{-2} \\\\
& x=4.6055
\end{aligned}

5) Solve -x^2+4x-20=0 using the quadratic formula.

x=-2, \quad x=6

x=-2i, \quad x=6i

x=2-4i, \quad x=2+4i

x=-2+4i \quad x=-2+4i

Identify a, b and c.

a=-1, \quad b=4, \quad c=-20

Substitute these values into the quadratic formula.

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-4 \pm \sqrt{(4)^2-4(-1)(-20)}}{2(-1)}
\end{aligned}

Solve the equation with a +, and then with a -.

\begin{aligned}
& x=\cfrac{-4+\sqrt{(4)^2-4(-1)(-20)}}{2(-1)} \\\\
& x=\cfrac{-4+\sqrt{16-80}}{-2} \\\\
& x=\cfrac{-4+\sqrt{64} i}{-2} \\\\
& x=2-4 i \end{aligned} \quad \begin{aligned}
& x=\cfrac{-4-\sqrt{(4)^2-4(-1)(-20)}}{2(-1)} \\\\
& x=\cfrac{-4-\sqrt{16-80}}{-2} \\\\
& x=\cfrac{-4-\sqrt{64} i}{-2} \\\\
& x=2+4 i
\end{aligned}

You can see that there are no values of x that give a y value of 0 .

The graph does not cross the x-axis. The use of the quadratic formula will result in roots that are not real.

6) Solve 3x^2-7x+21=0 using the quadratic formula.

x=-1.208i, \quad x=3.541i

x= \overline{1.16} + \sqrt{203}, \quad x=\overline{1.16} – \sqrt{203}

x=-1.208, \quad x=3.541

x=\overline{1.16}+2.3746i, \quad x=\overline{1.16}-2.3746i

Identify a, b and c.

a=3, \quad b=-7, \quad c=21

Substitute these values into the quadratic formula.

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-7) \pm \sqrt{(-7)^2-4(3)(21)}}{2(3)}
\end{aligned}

Solve the equation with a +, and then with a -.

\begin{aligned}
& x=\cfrac{-(-7)+\sqrt{(-7)^2-4(3)(21)}}{2(3)} \\\\
& x=\cfrac{7+\sqrt{49-252}}{6} \\\\
& x=\cfrac{7+\sqrt{203 i}}{6} \\\\
& x=1.1 \overline{6}+2.3746 i
\end{aligned} \quad \begin{aligned}
& x=\cfrac{-(-7)-\sqrt{(-7)^2-4(3)(21)}}{2(3)} \\\\
& x=\cfrac{7-\sqrt{49-252}}{6} \\\\
& x=\cfrac{7-\sqrt{203} i}{6} \\\\
& x=1.1 \overline{6}-2.3746 i
\end{aligned}

You can see that there are no values of x that give a y value of 0.

The graph does not cross the x -axis. The use of the quadratic formula will result in roots that are not real.

No, it only works for algebraic equations that have a second-degree term, otherwise known as quadratic equations. It will not work for polynomials of degrees other than two, such as linear (one-degree) or cubic equations (three-degree).

When the minimum (vertex) of the parabola formed by the quadratic equation is on the x -axis, it will only have one solution.

Yes, a can be positive or negative in a quadratic expression or equation. A negative a value causes the graph of the parabola to be flipped and open downward, instead of upward.

No, mathematicians have developed various ways to solve for the roots of a quadratic equation. Other ways include factoring the quadratic into the product of two terms or utilizing perfect squares to solve with a method called completing the square.

- Functions in algebra
- Laws of exponents
- Math formulas
- Algebraic fractions

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[FREE] Common Core Practice Tests (Grades 3 to 6)

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