Question

section 3.5 homework

1. solve the equation by graphing.

$x^2 - 3x + 1 = 2x^2 - 5x - 3$

a. $x \approx -7.13$ b. $x = 1$

c. $x \approx 2.36$ d. $x = -2$

e. $x \approx -1.24$ f. $x \approx 3.24$

g. $x = -3$ h. no real solution

Explanation:

Step1: Rearrange the equation

First, we need to rearrange the given equation \(x^{2}-3x + 1=2x^{2}-5x - 3\) to standard quadratic form. Subtract \(x^{2}-3x + 1\) from both sides:

$$
LATEXBLOCK0
$$

So the quadratic equation is \(x^{2}-2x - 4 = 0\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a
eq0\)), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b=-2\), \(c=-4\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times(-4)=4 + 16=20\).
Then, find the roots:

$$ x=\frac{-(-2)\pm\sqrt{20}}{2\times1}=\frac{2\pm2\sqrt{5}}{2}=1\pm\sqrt{5} $$

Calculate the numerical values: \(\sqrt{5}\approx2.236\), so \(1+\sqrt{5}\approx1 + 2.236=3.236\approx3.24\) and \(1-\sqrt{5}\approx1-2.236=-1.236\approx - 1.24\).

Step3: Match with the options

From the calculations, the solutions are approximately \(x\approx - 1.24\) and \(x\approx3.24\), which correspond to options e and f. But let's check the options again. Wait, maybe we made a mistake in rearrangement? Wait, let's re - check the original equation:

Original equation: \(x^{2}-3x + 1=2x^{2}-5x - 3\)

Bring all terms to one side: \(2x^{2}-5x - 3-(x^{2}-3x + 1)=0\)

\(2x^{2}-5x - 3 - x^{2}+3x - 1=0\)

\(x^{2}-2x - 4 = 0\), which is correct.

Now, let's check the options:

Option e: \(x\approx - 1.24\) (matches \(1-\sqrt{5}\approx - 1.24\))

Option f: \(x\approx3.24\) (matches \(1+\sqrt{5}\approx3.24\))

Wait, but let's check if we misread the options. The options are:

a. \(x\approx - 7.13\)

b. \(x = 1\)

c. \(x\approx2.36\)

d. \(x=-2\)

e. \(x\approx - 1.24\)

f. \(x\approx3.24\)

g. \(x=-3\)

h. no real solution

So the correct options are e and f. But maybe the question is a multiple - choice where we have to select the correct ones. Let's verify by plugging in the values.

For \(x\approx - 1.24\):

Left - hand side (LHS) of original equation: \((-1.24)^{2}-3\times(-1.24)+1=1.5376 + 3.72+1=6.2576\)

Right - hand side (RHS): \(2\times(-1.24)^{2}-5\times(-1.24)-3=2\times1.5376+6.2 - 3=3.0752 + 6.2-3=6.2752\approx6.2576\) (close, due to approximation)

For \(x\approx3.24\):

LHS: \((3.24)^{2}-3\times3.24 + 1=10.4976-9.72 + 1=1.7776\)

RHS: \(2\times(3.24)^{2}-5\times3.24-3=2\times10.4976-16.2 - 3=20.9952-16.2 - 3=1.7952\approx1.7776\) (close, due to approximation)

So the correct options are e. \(x\approx - 1.24\) and f. \(x\approx3.24\). But maybe the question is a single - choice? Wait, no, quadratic equations can have two solutions. But let's check the options again. Wait, maybe I made a mistake in the sign when moving terms. Let's try another approach: graph the two functions \(y = x^{2}-3x + 1\) and \(y=2x^{2}-5x - 3\). The x - coordinates of the intersection points are the solutions.

The vertex of \(y=x^{2}-3x + 1\): \(x=-\frac{b}{2a}=\frac{3}{2}=1.5\), \(y=(1.5)^{2}-3\times1.5 + 1=2.25-4.5 + 1=-1.25\)

The vertex of \(y = 2x^{2}-5x - 3\): \(x=\frac{5}{4}=1.25\), \(y=2\times(1.25)^{2}-5\times1.25-3=2\times1.5625-6.25 - 3=3.125-6.25 - 3=-6.125\)

By graphing (either by hand or using a graphing calculator), we can see that the two parabolas intersect at \(x\approx - 1.24\) and \(x\approx3.24\), which are options e and f.

Answer:

e. \(x\approx - 1.24\), f. \(x\approx3.24\)