Question

name
date
tracking a dive
diagnostic assessment
marcus and raj are discussing solutions to some equations.
marcus says that the solutions to $(x + 1)(x - 2)=0$ can be found by solving $x + 1 = 0$ and
$x - 2 = 0$ which means the solutions must be -1 or 2 in order for the product to equal 0.
raj says that the solutions to $x(x - 3)=7$ can be found by solving $x = 7$ and $x - 3 = 7$,
which means the solutions are 7 or 11 in order for the product to be 7.
who do you agree with? explain or justify your thinking.
name
date
lets solve some equations!

1. to solve the equation $n^{2}-2n = 99$, tyler wrote out the following steps. analyze tylers

work. write down what tyler did in each step.
$n^{2}-2n = 99$
original equation
$n^{2}-2n - 99 = 0$
step 1
$(n - 11)(n + 9)=0$
step 2
$n - 11 = 0$ or $n + 9 = 0$
step 3
$n = 11$ or $n=-9$
step 4

1. solve each equation either with tylers strategy or your own. show your reasoning.

$a. x^{2}+8x + 15 = 0$
$b. x^{2}-8x + 12 = 5$
$c. x^{2}-10x - 11 = 0$
$d. 49 - x^{2}=0$
$e. (x + 4)(x + 5)-30 = 0$

Explanation:

Step1: Analyze Marcus's statement

The zero - product property states that if \(ab = 0\), then \(a=0\) or \(b = 0\). For the equation \((x + 1)(x-2)=0\), by the zero - product property, we set \(x + 1=0\) (solving gives \(x=-1\)) and \(x - 2=0\) (solving gives \(x = 2\)).

Step2: Analyze Raj's statement

The zero - product property only applies when the product of two factors is equal to zero. For the equation \(x(x - 3)=7\), we cannot directly set \(x=7\) and \(x-3 = 7\) because the right - hand side is not zero. First, we need to expand and rewrite the equation in standard form: \(x^{2}-3x-7=0\) and then use methods like the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) (where \(a = 1\), \(b=-3\), \(c=-7\)) to solve it.

Answer:

I agree with Marcus. Marcus correctly applied the zero - product property (if \(ab = 0\), then \(a = 0\) or \(b=0\)). Raj incorrectly applied the zero - product property. The zero - product property is for equations of the form \(ab = 0\), not \(ab=c\) (\(c
eq0\)). Raj's equation \(x(x - 3)=7\) should be rewritten as \(x^{2}-3x - 7=0\) and solved using the quadratic formula \(x=\frac{3\pm\sqrt{(-3)^{2}-4\times1\times(-7)}}{2\times1}=\frac{3\pm\sqrt{9 + 28}}{2}=\frac{3\pm\sqrt{37}}{2}\), not by the method he used.