In previous lessons on quadratic equations and the discriminant, we learned how to find roots by factoring, completing the square, and using the quadratic formula. The roots of a quadratic equation may be different, equal, real, or imaginary.
Consider this problem. Suppose there are two consecutive even numbers whose product is 224. If the first number is x x x x + 2 x+2 x + 2
x ( x + 2 ) = 224
x(x+2)=224
x ( x + 2 ) = 224 or:
x 2 + 2 x − 224 = 0
x^2+2x-224=0
x 2 + 2 x − 224 = 0 Solving the equation gives x = 14 x=14 x = 14 14 and 16 .
Now, what if the roots of a quadratic equation are already known? How do we form the equation?
If a quadratic equation has roots x 1 x_1 x 1 x 2 x_2 x 2
( x − x 1 ) ( x − x 2 ) = 0
(x-x_1)(x-x_2)=0
( x − x 1 ) ( x − x 2 ) = 0 Determine the quadratic equation whose roots are:
− 2 -2 − 2 6 6 6 1 3 \dfrac{1}{3} 3 1 1 2 \dfrac{1}{2} 2 1 a. Roots − 2 -2 − 2 6 6 6 ( x − x 1 ) ( x − x 2 ) = 0 ( x − ( − 2 ) ) ( x − 6 ) = 0 ( x + 2 ) ( x − 6 ) = 0 x 2 − 6 x + 2 x − 12 = 0 x 2 − 4 x − 12 = 0
\begin{aligned}
(x-x_1)(x-x_2)&=0\\
(x-(-2))(x-6)&=0\\
(x+2)(x-6)&=0\\
x^2-6x+2x-12&=0\\
x^2-4x-12&=0
\end{aligned}
( x − x 1 ) ( x − x 2 ) ( x − ( − 2 )) ( x − 6 ) ( x + 2 ) ( x − 6 ) x 2 − 6 x + 2 x − 12 x 2 − 4 x − 12 = 0 = 0 = 0 = 0 = 0 So the required quadratic equation is:
x 2 − 4 x − 12 = 0
x^2-4x-12=0
x 2 − 4 x − 12 = 0 b. Roots 1 3 \dfrac{1}{3} 3 1 1 2 \dfrac{1}{2} 2 1 ( x − x 1 ) ( x − x 2 ) = 0 ( x − 1 3 ) ( x − 1 2 ) = 0 x 2 − 1 3 x − 1 2 x + 1 6 = 0 x 2 − 5 6 x + 1 6 = 0 6 x 2 − 5 x + 1 = 0
\begin{aligned}
(x-x_1)(x-x_2)&=0\\
\left(x-\frac{1}{3}\right)\left(x-\frac{1}{2}\right)&=0\\
x^2-\frac{1}{3}x-\frac{1}{2}x+\frac{1}{6}&=0\\
x^2-\frac{5}{6}x+\frac{1}{6}&=0\\
6x^2-5x+1&=0
\end{aligned}
( x − x 1 ) ( x − x 2 ) ( x − 3 1 ) ( x − 2 1 ) x 2 − 3 1 x − 2 1 x + 6 1 x 2 − 6 5 x + 6 1 6 x 2 − 5 x + 1 = 0 = 0 = 0 = 0 = 0 So the required quadratic equation is:
6 x 2 − 5 x + 1 = 0
6x^2-5x+1=0
6 x 2 − 5 x + 1 = 0 Quadratic Equations Discriminant Function Constructing a Quadratic Equation from the Sum and Product of Its Roots If the roots are x 1 x_1 x 1 x 2 x_2 x 2
( x − x 1 ) ( x − x 2 ) = 0 x 2 − x 1 x − x 2 x + x 1 x 2 = 0 x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0
\begin{aligned}
(x-x_1)(x-x_2)&=0\\
x^2-x_1x-x_2x+x_1x_2&=0\\
x^2-(x_1+x_2)x+x_1x_2&=0
\end{aligned}
( x − x 1 ) ( x − x 2 ) x 2 − x 1 x − x 2 x + x 1 x 2 x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0 = 0 = 0 Therefore, if the sum x 1 + x 2 x_1+x_2 x 1 + x 2 x 1 x 2 x_1x_2 x 1 x 2
x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0
x^2-(x_1+x_2)x+x_1x_2=0
x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0 From the quadratic formula:
x 1 = − b + b 2 − 4 a c 2 a and x 2 = − b − b 2 − 4 a c 2 a
x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}
\quad \text{and} \quad
x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}
x 1 = 2 a − b + b 2 − 4 a c and x 2 = 2 a − b − b 2 − 4 a c So:
x 1 + x 2 = − b a x 1 x 2 = c a
\begin{aligned}
x_1+x_2 &= \frac{-b}{a} \\
x_1x_2 &= \frac{c}{a}
\end{aligned}
x 1 + x 2 x 1 x 2 = a − b = a c Thus, for a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0
x 1 + x 2 = − b a and x 1 x 2 = c a
x_1+x_2=\frac{-b}{a}
\quad \text{and} \quad
x_1x_2=\frac{c}{a}
x 1 + x 2 = a − b and x 1 x 2 = a c Given:
x 1 + x 2 = 7 and x 1 x 2 = 10
x_1+x_2=7 \quad \text{and} \quad x_1x_2=10
x 1 + x 2 = 7 and x 1 x 2 = 10 Determine the quadratic equation.
x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0 x 2 − 7 x + 10 = 0
\begin{aligned}
x^2-(x_1+x_2)x+x_1x_2&=0\\
x^2-7x+10&=0
\end{aligned}
x 2 − ( x 1 + x 2 ) x + x 1 x 2 x 2 − 7 x + 10 = 0 = 0 If α \alpha α β \beta β
2 x 2 − 4 x − 12 = 0
2x^2-4x-12=0
2 x 2 − 4 x − 12 = 0 find:
α + β \alpha+\beta α + β α β \alpha\beta α β α β 2 + α 2 β \alpha\beta^2+\alpha^2\beta α β 2 + α 2 β α 2 + β 2 \alpha^2+\beta^2 α 2 + β 2 α + β = − b a = − ( − 4 ) 2 = 2
\begin{aligned}
\alpha+\beta&=\frac{-b}{a}
=\frac{-(-4)}{2}
=2
\end{aligned}
α + β = a − b = 2 − ( − 4 ) = 2 α β = c a = − 12 2 = − 6
\begin{aligned}
\alpha\beta&=\frac{c}{a}
=\frac{-12}{2}
=-6
\end{aligned}
α β = a c = 2 − 12 = − 6 α β 2 + α 2 β = α β ( α + β ) = ( − 6 ) ( 2 ) = − 12
\begin{aligned}
\alpha\beta^2+\alpha^2\beta
&=\alpha\beta(\alpha+\beta)\\
&=(-6)(2)\\
&=-12
\end{aligned}
α β 2 + α 2 β = α β ( α + β ) = ( − 6 ) ( 2 ) = − 12 α 2 + β 2 = ( α + β ) 2 − 2 α β = 2 2 − 2 ( − 6 ) = 16
\begin{aligned}
\alpha^2+\beta^2
&=(\alpha+\beta)^2-2\alpha\beta\\
&=2^2-2(-6)\\
&=16
\end{aligned}
α 2 + β 2 = ( α + β ) 2 − 2 α β = 2 2 − 2 ( − 6 ) = 16 If α \alpha α β \beta β
2 x 2 − 14 x + 12 = 0
2x^2-14x+12=0
2 x 2 − 14 x + 12 = 0 determine a new quadratic equation whose roots are α + 2 \alpha+2 α + 2 β + 2 \beta+2 β + 2
From 2 x 2 − 14 x + 12 = 0 2x^2-14x+12=0 2 x 2 − 14 x + 12 = 0
a = 2 , b = − 14 , c = 12
a=2,\quad b=-14,\quad c=12
a = 2 , b = − 14 , c = 12 α + β = − b a = 14 2 = 7
\alpha+\beta=\frac{-b}{a}=\frac{14}{2}=7
α + β = a − b = 2 14 = 7 α β = c a = 12 2 = 6
\alpha\beta=\frac{c}{a}=\frac{12}{2}=6
α β = a c = 2 12 = 6 The new sum is:
( α + 2 ) + ( β + 2 ) = 7 + 4 = 11
(\alpha+2)+(\beta+2)=7+4=11
( α + 2 ) + ( β + 2 ) = 7 + 4 = 11 The new product is:
( α + 2 ) ( β + 2 ) = 6 + 2 ( 7 ) + 4 = 24
(\alpha+2)(\beta+2)=6+2(7)+4=24
( α + 2 ) ( β + 2 ) = 6 + 2 ( 7 ) + 4 = 24 So the new quadratic equation is:
x 2 − [ ( α + 2 ) + ( β + 2 ) ] x + ( α + 2 ) ( β + 2 ) = 0 x 2 − 11 x + 24 = 0
\begin{aligned}
x^2-[(\alpha+2)+(\beta+2)]x+(\alpha+2)(\beta+2)&=0\\
x^2-11x+24&=0
\end{aligned}
x 2 − [( α + 2 ) + ( β + 2 )] x + ( α + 2 ) ( β + 2 ) x 2 − 11 x + 24 = 0 = 0 Many daily-life problems can be modeled using quadratic equations. The general steps are:
let one unknown be x x x form an equation in x x x solve the equation, choose the solution that fits the context. The difference between two whole numbers is 2 and their product is 168. Determine the two numbers.
Let the first number be x x x x − 2 x-2 x − 2
x ( x − 2 ) = 168 x 2 − 2 x = 168 x 2 − 2 x − 168 = 0 ( x + 12 ) ( x − 14 ) = 0 x = − 12 or x = 14
\begin{aligned}
x(x-2)&=168\\
x^2-2x&=168\\
x^2-2x-168&=0\\
(x+12)(x-14)&=0\\
x&=-12 \quad \text{or} \quad x=14
\end{aligned}
x ( x − 2 ) x 2 − 2 x x 2 − 2 x − 168 ( x + 12 ) ( x − 14 ) x = 168 = 168 = 0 = 0 = − 12 or x = 14 Since − 12 -12 − 12 x = 14 x=14 x = 14
Therefore, the two numbers are 12 and 14 .
Mr. Harun’s garden is rectangular. Its length is 3 meters more than its width. If the area is 108 m 2 108 \text{ m}^2 108 m 2
Let the length be p p p p − 3 p-3 p − 3
108 = p ( p − 3 ) 108 = p 2 − 3 p p 2 − 3 p − 108 = 0 ( p − 12 ) ( p + 9 ) = 0
\begin{aligned}
108&=p(p-3)\\
108&=p^2-3p\\
p^2-3p-108&=0\\
(p-12)(p+9)&=0
\end{aligned}
108 108 p 2 − 3 p − 108 ( p − 12 ) ( p + 9 ) = p ( p − 3 ) = p 2 − 3 p = 0 = 0 So:
p = 12 or p = − 9
p=12 \quad \text{or} \quad p=-9
p = 12 or p = − 9 Only p = 12 p=12 p = 12
l = 12 − 3 = 9
l=12-3=9
l = 12 − 3 = 9 The perimeter is:
K = 2 ( p + l ) = 2 ( 12 + 9 ) = 42
\begin{aligned}
K&=2(p+l)\\
&=2(12+9)\\
&=42
\end{aligned}
K = 2 ( p + l ) = 2 ( 12 + 9 ) = 42 So the perimeter of Mr. Harun’s garden is 42 meters .
After studying quadratic equations, the discriminant function, and composing quadratic equations, continue to the evaluation below:
Quadratic Equation Quiz
That concludes this material on composing quadratic equations for Grade IX SMP/MTs.
