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 $y = mx + c$.

Step 1: Calculate the slope ($m$) of the line using the two given points A$(x_1, y_1) = (2, 4)$ and B$(x_2, y_2) = (5, 13)$. $m = \frac{y_2 - y_1}{x_2 - x_1}$ $m = \frac{13 - 4}{5 - 2}$ $m = \frac{9}{3}$ $m = 3$

Step 2: Use the slope $m=3$ and one of the points (e.g., A(2, 4)) to find the y-intercept ($c$) using the equation $y = mx + c$. $4 = 3(2) + c$ $4 = 6 + c$ $c = 4 - 6$ $c = -2$

Step 3: 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 $\sqrt{x+2} = x$.

Step 1: Square both sides of the equation to eliminate the square root. $(\sqrt{x+2})^2 = x^2$ $x+2 = x^2$

Step 2: Rearrange the equation into a standard quadratic form $ax^2 + bx + c = 0$. $x^2 - x - 2 = 0$

Step 3: Solve the quadratic equation by factoring. $(x - 2)(x + 1) = 0$ This gives two possible solutions: $x - 2 = 0 \implies x = 2$ $x + 1 = 0 \implies x = -1$

Step 4: 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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