alain throws a stone off a bridge into a river below. the stones height (in meters above the water), $x$…

alain throws a stone off a bridge into a river below. the stones height (in meters above the water), $x$ seconds after alain threw it, is modeled by: $h(x)=-5x^{2}+10x + 15$. how many seconds after being thrown will the stone hit the water? seconds

alain throws a stone off a bridge into a river below. the stones height (in meters above the water), $x$ seconds after alain threw it, is modeled by: $h(x)=-5x^{2}+10x + 15$. how many seconds after being thrown will the stone hit the water? seconds

Answer

Explanation:

Step1: Set height to 0

When the stone hits the water, $h(x)=0$. So we set $-5x^{2}+10x + 15=0$.

Step2: Divide by - 5

Divide the entire equation by - 5 to simplify. We get $x^{2}-2x - 3=0$.

Step3: Factor the quadratic

Factor $x^{2}-2x - 3$ as $(x - 3)(x+1)=0$.

Step4: Solve for x

Using the zero - product property, if $(x - 3)(x + 1)=0$, then $x-3=0$ or $x + 1=0$. So $x=3$ or $x=-1$. Since time $x$ cannot be negative in this context, we discard $x=-1$.

Answer:

3