Find the equation passing through A(2, 4) and B(5, 13) in form y = mx + c. Solve the equation √x + 2 = x2

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Find the equation passing through A(2, 4) and B(5, 13) in form $y = mx + c$ .

Calculate the slope ( $m$ ) of the line using the two given points A $(x_1, y_1) = (2, 4)$ and B $(x_2, y_2) = (5, 13)$ .
$m = \frac{y_2 - y_1}{x_2 - x_1}$ $m = \frac{13 - 4}{5 - 2}$ $m = \frac{9}{3}$ $m = 3$
Use the slope $m=3$ and one of the points (e.g., A(2, 4)) to find the y-intercept ( $c$ ) using the equation $y = mx + c$ .
$4 = 3(2) + c$ $4 = 6 + c$ $c = 4 - 6$ $c = -2$
Write the equation of the line in the form $y = mx + c$ .
$y = 3x - 2$
The equation of the line is $\boxed{y = 3x - 2}$ .

Solve the equation $\sqrt{x+2} = x$ .
Square both sides of the equation to eliminate the square root.
$(\sqrt{x+2})^2 = x^2$ $x+2 = x^2$
Rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$ .
$x^2 - x - 2 = 0$
Solve the quadratic equation by factoring.
$(x - 2)(x + 1) = 0$
This gives two possible solutions:
$x - 2 = 0 \implies x = 2$ $x + 1 = 0 \implies x = -1$
Check both possible solutions in the original equation $\sqrt{x+2} = x$ to identify any extraneous solutions.
For $x = 2$ :
$\sqrt{2+2} = \sqrt{4} = 2$ $2 = 2$
This solution is valid.

For $x = -1$ :
$\sqrt{-1+2} = \sqrt{1} = 1$ $1 \neq -1$
This solution is extraneous because the square root symbol $\sqrt{}$ denotes the principal (non-negative) square root.

The only valid solution is $\boxed{x = 2}$ .

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