Question

1. when drawn on the same grid, the graphs of $y = x$ and $y = \sqrt{2x + 3}$ will intersect at

(1) $(-1, 1)$
(3) $(3, 3)$
(2) $(2, \frac{1}{2})$
(4) $(-4, -4)$
\\(\sqrt{2x + 3} = x\\)
algebra 2 with trigonometry, unit reviews - unit #4
emathinstruction, red hook, ny 12571, \\(\odot\\) 2009

Explanation:

Step1: Set the equations equal

To find the intersection, set \( y = x \) and \( y=\sqrt{2x + 3} \) equal: \( x=\sqrt{2x + 3} \)

Step2: Square both sides

Square both sides to eliminate the square root: \( x^{2}=2x + 3 \)

Step3: Rearrange into quadratic

Rearrange to \( x^{2}-2x - 3 = 0 \)

Step4: Factor the quadratic

Factor: \( (x - 3)(x+ 1)=0 \)

Step5: Solve for x

Solutions: \( x = 3 \) or \( x=-1 \)

Step6: Check solutions

• For \( x=-1 \): \( y=-1 \), but \( \sqrt{2(-1)+3}=\sqrt{1} = 1

eq - 1 \), so extraneous.

• For \( x = 3 \): \( y = 3 \), and \( \sqrt{2(3)+3}=\sqrt{9}=3 \), valid.

Answer:

(3) (3, 3)