Question

solve using factoring
$x^2 - 4x - 17 = 4$
select the correct response:
13, -3
-7, 3
-7
-3, 7

Explanation:

Step1: Rearrange the equation

First, we need to set the equation to zero. Subtract 4 from both sides of the equation \(x^{2}-4x - 17=4\).

$$x^{2}-4x-17 - 4=0$$

$$x^{2}-4x - 21=0$$

Step2: Factor the quadratic equation

We need to find two numbers that multiply to \(- 21\) and add up to \(-4\). The numbers are \(-7\) and \(3\) since \((-7)\times3=-21\) and \(-7 + 3=-4\).
So we can factor the quadratic as:

$$(x - 7)(x+3)=0$$

Wait, no, let's correct that. Wait, the middle term is \(-4x\), so the correct factoring is \((x - 7)(x + 3)=x^{2}+3x-7x - 21=x^{2}-4x - 21\), yes. Wait, actually, it's \((x - 7)(x+3)=0\)? Wait, no, if we have \(x^{2}-4x - 21=(x - 7)(x + 3)\)? Wait, \(x\times x=x^{2}\), \(x\times3=3x\), \(-7\times x=-7x\), \(-7\times3 = - 21\), then \(3x-7x=-4x\), yes. So the factored form is \((x - 7)(x + 3)=0\)

Step3: Solve for x

Set each factor equal to zero:
Case 1: \(x - 7=0\), then \(x = 7\)? Wait, no, wait, that's a mistake. Wait, no, the factoring should be \((x - 7)(x+3)=0\)? Wait, no, let's check again. Wait, we need two numbers that multiply to \(-21\) and add to \(-4\). Let's list the factors of \(-21\): \((-21)\) and \(1\) (sum \(-20\)), \((-7)\) and \(3\) (sum \(-4\)), \(7\) and \(-3\) (sum \(4\)), \(21\) and \(-1\) (sum \(20\)). So the correct factoring is \(x^{2}-4x - 21=(x - 7)(x + 3)\)? Wait, no, if we have \(x^{2}-4x - 21\), then when we factor, it's \((x - 7)(x + 3)\) because \(x\times x=x^{2}\), \(x\times3 = 3x\), \(-7\times x=-7x\), \(-7\times3=-21\), and \(3x-7x=-4x\). So setting \((x - 7)(x + 3)=0\), then \(x - 7=0\) gives \(x = 7\) and \(x+3=0\) gives \(x=-3\)? Wait, but the options are \(13,-3\); \(-7,3\); \(-7\); \(-3,7\). Wait, I must have made a mistake in factoring. Wait, let's do it again. Wait, the equation is \(x^{2}-4x - 21=0\). Wait, maybe I mixed up the signs. Let's use the quadratic formula to check. The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b=-4\), \(c=-21\). Then \(b^{2}-4ac=(-4)^{2}-4\times1\times(-21)=16 + 84=100\). Then \(\sqrt{100}=10\). So \(x=\frac{4\pm10}{2}\). So \(x=\frac{4 + 10}{2}=\frac{14}{2}=7\), and \(x=\frac{4-10}{2}=\frac{-6}{2}=-3\). Wait, but the options have \(-3,7\) as one of the options (the last option: \(-3,7\)). Wait, maybe I made a mistake in the factoring earlier. Let's re - factor. \(x^{2}-4x - 21\). We need two numbers that multiply to \(-21\) and add to \(-4\). The numbers are \(-7\) and \(3\)? Wait, no, \(-7+3=-4\), but when we multiply \(-7\times3=-21\). So the factoring is \(x^{2}-4x - 21=(x - 7)(x + 3)\)? Wait, but when we expand \((x - 7)(x + 3)=x^{2}+3x-7x - 21=x^{2}-4x - 21\), correct. Then setting \((x - 7)(x + 3)=0\) gives \(x = 7\) or \(x=-3\), which is the option \(-3,7\) (the last option).

Wait, maybe I had a sign error in the factoring step earlier when I thought \(x - 7 = 0\) gives \(x = 7\), which is correct, and \(x+3=0\) gives \(x=-3\), so the solutions are \(x = 7\) and \(x=-3\), which is the option \(-3,7\) (the last option: \(-3,7\))

Answer:

\(-3,7\) (the option with \(-3\) and \(7\))