The path of a fired projectile is modeled by the function f(x)=x(2x+8) where x is the height (in feet) and f(x) in the distance(in feet). If the distance is 24 feet, what is the height of the projectile?
+26367 The path of a fired projectile is modeled by the function f(x)=x(2x+8) where x is the height (in feet) and f(x) in the distance(in feet). If the distance is 24 feet, what is the height of the projectile ?
$$f(x)=x(2x+8)\quad f(x) = 24 \\
24 = x(2x+8) \\
24 = 2x^2+8x \quad | \quad :2\\
12 = x^2 + 4x \quad | \quad -12\\
0 = x^2 + 4x -12 \\
x^2 + 4x -12 = 0 \quad \small{\text{factoring}}\\
\underbrace{(x-2)}_{=0}\underbrace{(x+6)}_{=0} =0 \\
x-2=0 \quad \small{\text{we have} }
\quad x=2 \quad \small{\text{okay}} \\
x+6=0 \quad \small{\text{we have}} \quad x=-6 \quad \small{\text{no solution!}}\\$$
the height of the projectile is 2 feet.
+26367 The path of a fired projectile is modeled by the function f(x)=x(2x+8) where x is the height (in feet) and f(x) in the distance(in feet). If the distance is 24 feet, what is the height of the projectile ?
$$f(x)=x(2x+8)\quad f(x) = 24 \\
24 = x(2x+8) \\
24 = 2x^2+8x \quad | \quad :2\\
12 = x^2 + 4x \quad | \quad -12\\
0 = x^2 + 4x -12 \\
x^2 + 4x -12 = 0 \quad \small{\text{factoring}}\\
\underbrace{(x-2)}_{=0}\underbrace{(x+6)}_{=0} =0 \\
x-2=0 \quad \small{\text{we have} }
\quad x=2 \quad \small{\text{okay}} \\
x+6=0 \quad \small{\text{we have}} \quad x=-6 \quad \small{\text{no solution!}}\\$$
the height of the projectile is 2 feet.
