Question

name: zoey chapman
date: _____________ per: ____
** this is a 2 - page d
directions: find the solutions of each quadratic equ

1. $x^2 - 4x = 0$

2.

1. $x^2 - 36 = 0$

4.

1. $x^2 + 12x + 27 = 0$

6.

1. $x^2 = x + 30$

8

1. $2x^2 + 6x - 80 = 0$

1

Explanation:

Problem 1: \( x^2 - 4x = 0 \)

Step 1: Factor the equation

Factor out \( x \) from the left - hand side: \( x(x - 4)=0 \)

Step 2: Apply the zero - product property

If \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So we have two cases:
Case 1: \( x=0 \)
Case 2: \( x - 4=0 \), which gives \( x = 4 \)

Step 1: Recognize the difference of squares

We know that \( a^{2}-b^{2}=(a + b)(a - b) \). Here, \( a=x \) and \( b = 6 \), so \( x^{2}-36=(x + 6)(x - 6)=0 \)

Step 2: Apply the zero - product property

Set each factor equal to zero:
Case 1: \( x+6 = 0\Rightarrow x=-6 \)
Case 2: \( x - 6=0\Rightarrow x = 6 \)

Step 1: Factor the quadratic

We need to find two numbers that multiply to \( 27 \) and add up to \( 12 \). The numbers are \( 9 \) and \( 3 \). So \( x^{2}+12x + 27=(x + 9)(x+3)=0 \)

Step 2: Apply the zero - product property

Set each factor equal to zero:
Case 1: \( x + 9=0\Rightarrow x=-9 \)
Case 2: \( x + 3=0\Rightarrow x=-3 \)

Answer:

\( x = 0 \) or \( x = 4 \)

Problem 3: \( x^2-36 = 0 \)