Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 6.2
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Exercise 84 Page 353

Practice makes perfect
a To solve this equation we will apply the Zero Product Property.
(5x-1)(x+3)=0
lc5x-1=0 & (I) x+3=0 & (II)
l5x=1 x+3=0
lx= 15 x+3=0
lx_1= 15 x_2=- 3
b We want to solve the given equation by factoring.

Factoring

This equation would be much easier to factor if it had smaller coefficients. We can start factoring by changing this equation to a simpler equivalent equation by dividing both sides of the equation by 2.
4x^2+10x-6=0
2x^2+5x-3=0
2x^2+6x-x-3=0
â–Ľ
Factor out 2x & - 1
2x(x+3)-x-3=0
2x(x+3)-1(x+3)=0
(2x-1)(x+3)=0

Solving

To solve this equation, we will apply the Zero Product Property.
(2x-1)(x+3)=0
lc2x-1=0 & (I) x+3=0 & (II)
â–Ľ
(I), (II): Solve for x
l2x=1 x+3=0
lx= 12 x+3=0
lx_1= 12 x_2=- 3
c We will use the Quadratic Formula to solve the given quadratic equation.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aWe first need to identify the values of a, b, and c. x^2+2x+5=0 ⇕ 1x^2+ 2x+ 5=0 We see that a= 1, b= 2, and c= 5. Let's substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 2±sqrt(2^2-4( 1)( 5))/2( 1)
â–Ľ
Solve for x and Simplify
x=- 2±sqrt(4-4(1)(5))/2(1)
x=- 2±sqrt(4-4(5))/2
x=- 2±sqrt(4-20)/2
x=- 2±sqrt(- 16)/2 *
Oops! The square root of a negative number is undefined for real numbers! As we can see, using the Quadratic Formula for the given equation resulted in contradiction. Thus, there are no real solutions. For further explanation, notice that the left-hand side of the equation is a quadratic function. Let's graph it.

The solutions of the given equation are x-intercepts of this function. Notice that this parabola lies above the x-axis, so it does not have x-intercepts. Therefore there are no real solutions of the given equation.

d We want to solve the given equation by factoring.

Factoring

Let's start by writing all the terms on one side of the equals sign.
x^2+5x=14
x^2+5x-14=0
x^2+7x-2x-14=0
â–Ľ
Factor out x & - 2
x(x+7)-2x-14=0
x(x+7)-2(x+7)=0
(x-2)(x+7)=0

Solving

To solve this equation we will apply the Zero Product Property.
(x-2)(x+7)=0
lcx-2=0 & (I) x+7=0 & (II)
lx=2 x+7=0
lx_1=2 x_2=- 7