Question

Choose a method (factoring. square root property, completing the square, and the quadratic formula) to solve each of the following quadratic equations.
Each method can be used only ONCE

A) 4x2 – 27 = 0

B) 4x2 – 8x – 5 = 0

C) 4x2 – 8x – 12 = 0

D) 4x2 – 9x – 7 = 0

Answer 1

Answer:

Part A) $x=(+/-)\frac{3\sqrt{3}}{2}$

Part B) $x=2.5$  and $x=-0.5$

Part C) $x=-1$  and $x=3$

Part D) $x=\frac{9+\sqrt{193}}{8}$  and $x=\frac{9-\sqrt{193}}{8}$

Step-by-step explanation:

Part A) we have

$4x^{2}-27=0$          

we know that      

The square root property states that if we have an equation with a perfect square on one side and a number on the other side, then we can take the square root of both sides and add a plus or minus sign to the side with the number and solve the equation.

isolate the term that contains the squared variable

$4x^{2}=27$            

$x^{2} =\frac{27}{4}$

take the square root of both sides

$x=(+/-)\frac{\sqrt{27}}{2}$

simplify

$x=(+/-)\frac{3\sqrt{3}}{2}$

Part B) we have

$4x^{2}-8x-5=0$    

Using the quadratic equation  

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$ is  

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^{2}-8x-5=0$  

so

$a=4\\b=-8\\c=-5$

substitute in the formula

$x=\frac{-(-8)(+/-)\sqrt{-8^{2}-4(4)(-5)}} {2(4)}$

$x=\frac{8(+/-)\sqrt{144}}{8}$

$x=\frac{8(+/-)12}{8}$

$x=\frac{8(+)12}{8}=2.5$

$x=\frac{8(-)12}{8}=-0.5$

Part C) we have

$4x^{2}-8x-12=0$    

Using Factoring  

Simplify the expression first

Divide by 4 both sides

$x^{2}-2x-3=0$  

Find two numbers a and  b such that

a+b=-2

ab=-3

Solve the system by graphing

The solution is a=1, b=-3  

see the attached figure

so

$4x^{2}-8x-12=4(x-1)(x+3)$  

The solutions are

$x=1, x=-3$

Part D) we have

$4x^{2}-9x-7=0$

Solve by completing the square

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$4x^{2}-9x=7$

Factor the leading coefficient

$4(x^{2}-\frac{9}{4}x)=7$

Complete the square. Remember to balance the equation by adding the same constants to each side

$4(x^{2}-\frac{9}{4}x+\frac{81}{64})=7+\frac{81}{16}$

$4(x^{2}-\frac{9}{4}x+\frac{81}{64})=\frac{193}{16}$

Rewrite as perfect squares

$4(x-\frac{9}{8})^{2}=\frac{193}{16}$

$(x-\frac{9}{8})^{2}=\frac{193}{64}$

take square root both sides

$(x-\frac{9}{8})=(+/-)\frac{\sqrt{193}}{8}$

$x=\frac{9+\sqrt{193}}{8}$

$x=\frac{9-\sqrt{193}}{8}$