Question

select the correct answer.
the equations $y = 2x - 5$ and $y = \sqrt{3x - 1}$ are graphed on the coordinate grid.
how many real solutions does the equation $\sqrt{3x - 1} = 2x - 5$ have?
a. 0
b. 1
c. 2
d. cannot be determined from the graph

Explanation:

Step1: Set equations equal, square both sides

$$\sqrt{3x - 1} = 2x - 5$$
Square both sides:
$$3x - 1 = (2x - 5)^2$$

Step2: Expand right-hand side

$$3x - 1 = 4x^2 - 20x + 25$$

Step3: Rearrange to quadratic form

$$4x^2 - 23x + 26 = 0$$

Step4: Solve quadratic equation

Use quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=4$, $b=-23$, $c=26$:
$$x = \frac{23 \pm \sqrt{(-23)^2 - 4(4)(26)}}{2(4)}$$
$$x = \frac{23 \pm \sqrt{529 - 416}}{8}$$
$$x = \frac{23 \pm \sqrt{113}}{8}$$
Calculate approximate values:
$x \approx \frac{23 + 10.63}{8} \approx 4.20$; $x \approx \frac{23 - 10.63}{8} \approx 1.55$

Step5: Check for extraneous solutions

For $x \approx 1.55$:
Right-hand side: $2(1.55) - 5 = 3.1 - 5 = -1.9$
Left-hand side: $\sqrt{3(1.55)-1} = \sqrt{3.65} \approx 1.91$
$-1.9
eq 1.91$, so this is extraneous.

For $x \approx 4.20$:
Right-hand side: $2(4.20) - 5 = 8.4 - 5 = 3.4$
Left-hand side: $\sqrt{3(4.20)-1} = \sqrt{11.6} \approx 3.41$
These are approximately equal, so this is a valid solution.

Also, note the domain of $\sqrt{3x-1}$ requires $3x-1 \geq 0 \implies x \geq \frac{1}{3}$, and $2x-5 \geq 0$ (since square root is non-negative) $\implies x \geq 2.5$. The extraneous solution $x \approx 1.55$ does not satisfy $x \geq 2.5$.

Answer:

B. 1