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College Algebra/Solving Quadratic Equations and Inequalities/Quadratic Formula

Solving Quadratic Equations and Inequalities

Contents

Quadratic Formula

Deriving the Quadratic Formula

Completing the square for the standard form of a quadratic equation results in the quadratic formula.
The quadratic formula is used to solve a quadratic equation of the form ax2+bx+c=0ax^2+bx+c=0, where a≠0a\neq0, by using its coefficients:
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
The quadratic formula can be used to solve any quadratic equation. It is found by completing the square for the standard form of a quadratic equation.

How to Derive the Quadratic Formula

Standard Form Steps
ax2+bx+c=0ax^2+bx+c=0, a≠0a\neq0 The equation is in standard form.
aax2+bax+ca=0ax2+bax+ca=0\begin{aligned}\frac{a}{a}x^2+\frac{b}{a}x+\frac{c}{a}&=\frac{0}{a}\\x^2+\frac{b}{a}x+\frac{c}{a}&=0\end{aligned}
 Divide both sides by aa.
x2+bax=−cax^2+\frac{b}{a}x=-\frac{c}{a}
 Subtract ca\frac{c}{a} from both sides.
x2+bax+(b2a)2=−ca+(b2a)2x^2+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^{2}=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^{2}
 Add (b2a)2\left ( \frac{b}{2a} \right )^{2} to the left side to complete the square. Add (b2a)2\left ( \frac{b}{2a} \right )^{2} to the right side to keep the equation balanced.
(x+b2a)2=b24a2−ca\left (x+\frac{b}{2a}\right )^2=\frac{b^2}{4a^2}-\frac{c}{a}
 Factor and simplify.
(x+b2a)2=b24a2−4ac4a2(x+b2a)2=b2−4ac4a2\begin{aligned}\left (x+\frac{b}{2a}\right )^2&=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\\ \left (x+\frac{b}{2a}\right )^2&=\frac{b^2-4ac}{4a^2}\end{aligned}
 Use a common denominator on the right side, and simplify.
x+b2a=±b2−4ac4a2x+b2a=±b2−4ac2a\begin{aligned}x+\frac{b}{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\x+\frac{b}{2a}&=\pm\frac{\sqrt{b^2-4ac}}{2a}\end{aligned}
 Take the square root of both sides, and simplify.
x=−b2a±b2−4ac2ax=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}
 Subtract b2a\frac{b}{2a} from both sides.
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
 Simplify.

Using the Quadratic Formula

The quadratic formula can be applied to solve any quadratic equation. The discriminant of the related quadratic equation can be used to determine the number and type of roots of the quadratic function.
The discriminant is the part of the quadratic formula that is under the radical sign:
b2−4acb^2-4ac
Recall that a quadratic equation may have zero, one, or two real solutions. The value of the discriminant determines how many real solutions a quadratic equation has:
  • If b2−4acb^2-4ac is positive, then the equation has two real solutions.
  • If b2−4acb^2-4ac is zero, then the equation has one real solution.
  • If b2−4acb^2-4ac is negative, then the equation has zero real solutions.
Step-By-Step Example
Using the Quadratic Formula with Two Real Solutions
Solve the quadratic equation:
x2−4x−6=0x^2-4x-6=0
Step 1
The equation is in the standard form, with a=1a=1, b=−4b=-4, and c=−6c=-6:
ax2+bx+c=0x2−4x−6=0\begin{aligned}ax^2+bx+c&=0\\x^2-4x-6&=0\end{aligned}
The discriminant from the quadratic formula is the expression under the radical symbol:
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}
Use the discriminant from the quadratic formula to determine the number of solutions:
b2−4ac(−4)2−4(1)(−6)16+2440\begin{gathered}b^2-4ac\\(-4)^2-4(1)(-6)\\16+24\\40\end{gathered}
The discriminant is positive. So, the equation has two real solutions.
Step 2
Using the quadratic formula, substitute the value of the discriminant under the radical. Then substitute the values for aa and bb.
x=−b±b2−4ac2ax=−(−4)±402(1)\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-(-4)\pm\sqrt{40}}{2(1)}\end{aligned}
Step 3
Simplify the expression.
x=4±402x=4±2102x=2±10\begin{aligned}x&=\frac{4\pm\sqrt{40}}{2}\\x&=\frac{4\pm2\sqrt{10}}{2}\\x&=2\pm\sqrt{10}\end{aligned}
Solution
The two solutions of the equation are:
x=2+10orx=2−10x=2+\sqrt{10}\;\;\;\;\;\text{or}\;\;\;\;\;x=2-\sqrt{10}
Step-By-Step Example
Using the Quadratic Formula with One Real Solution
Solve the quadratic formula:
2x2+8x+8=02x^2+8x+8=0
Step 1
The equation is in the standard form, with a=2a=2, b=8b=8, and c=8c=8.
ax2+bx+c=02x2+8x+8=0\begin{aligned}ax^2+bx+c&=0\\2x^2+8x+8&=0\end{aligned}
The discriminant from the quadratic formula is the expression under the radical symbol:
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}
Use the discriminant from the quadratic formula to determine the number of solutions:
b2−4ac(8)2−4(2)(8)64−640\begin{gathered}b^2-4ac\\(8)^2-4(2)(8)\\64-64\\0\end{gathered}
The discriminant is zero. So, the equation has 1 real solution.
Step 2
Using the quadratic formula, substitute the value of the discriminant under the radical. Then substitute the values for aa and bb.
x=−b±b2−4ac2ax=−8±02(2)\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-8\pm\sqrt{0}}{2(2)}\end{aligned}
Solution
Simplify the expression.
x=−8±04x=−84x=−2\begin{aligned}x&=\frac{-8\pm0}{4}\\x&=\frac{-8}{4}\\x&=-2\end{aligned}
The solution is x=−2x=-2.
Step-By-Step Example
Using the Quadratic Formula with No Real Solutions
Solve the quadratic equation:
x2−6x+11=0x^2-6x+11=0
Step 1
The equation is in the standard form, with a=1a=1, b=−6b=-6, and c=11c=11.
ax2+bx+c=0x2−6x+11=0\begin{aligned}ax^2+bx+c&=0\\x^2-6x+11&=0\end{aligned}
The discriminant from the quadratic formula is the expression under the radical symbol:
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}
Use the discriminant from the quadratic formula to determine the number of solutions:
b2−4ac(−6)2−4(1)(11)\begin{gathered}b^2-4ac\\(-6)^2-4(1)(11)\end{gathered}
Step 2
Evaluate the discriminant.
(−6)2−4(1)(11)(−6)(−6)−(4)(11)36−44−8\begin{gathered}(-6)^2-4(1)(11)\\(-6)(-6)-(4)(11)\\36-44\\-8\end{gathered}
Solution
The discriminant is negative. So, the equation has zero real solutions. There is no need to use the quadratic formula.
<Completing the Square>Solving Quadratic Equations