Question Video: Creating and Using Quadratic Equations in More Than One Variable Mathematics • 10th Grade

In a baseball practice, Jacob threw the ball such that ℎ, its height above the ground in feet 𝑡 seconds after he released it, could be modelled by the quadratic equation ℎ = −16𝑡² + 46𝑡 + 5. Write an equation that could be solved to find the times at which the ball was at a height of 33 feet above the ground. For how long was the ball 33 feet above the ground?

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Video Transcript

In a baseball practice, Jacob threw the ball such that ℎ, its height above the ground in feet 𝑡 seconds after he released it, could be modelled by the quadratic equation ℎ equals negative 16 𝑡 squared plus 46 𝑡 plus five. Write an equation that could be solved to find the times at which the ball was at a height of 33 feet above the ground. For how long was the ball 33 feet above the ground?

Okay, so this question actually has two clear parts that needs to be solved. The first part is to write an equation that could be solved to find the times at which the ball was at the height of 33 feet above the ground. But a key bit of information here is the fact that the height was 33 feet above the ground. Because if we actually take a look back at the first part of the question, it tells us that ℎ is the height above the ground in feet.

So great, okay, so we now know that the height is 33 feet above the ground. So therefore, we can substitute the value ℎ is equal to 33 into our quadratic equation. So therefore, if we do, we’re actually gonna get that negative 16 𝑡 squared plus 46 𝑡 plus five is equal to 33. So great, we’ve actually solved the first part of the question.

Now, we’ve got to move on to the second part. And the second part it says, for how long was the ball 33 feet above the ground? Now, to actually answer the second part of the question, what we’re gonna need to do is actually we need to find the times when the ball was actually 33 feet above the ground. The first thing we will do to enable me to solve this equation to find 𝑡 is actually rearrange it. And the first thing we’re gonna do is subtract five from each side. So we get negative 16 𝑡 squared plus 46 𝑡 equals 28.

So now, if we want to solve a quadratic, we want it equal to zero. And to enable that to happen, what we’ve done is we’ve added 16 𝑡 squared and subtracted 46 𝑡 from each side. Therefore, we’re left with the equation zero is equal to 16 𝑡 squared minus 46 𝑡 plus 28. Okay, great, we’ve got our equation that we want to solve. So let’s go ahead and solve it.

Okay, so we’ve just rewritten our equation, just with the zero on the right-hand side. And now, the first thing we’re gonna do so in the steps to solve it is to actually divide three by two cause it would just make all the values more manageable. So this is gonna give us eight 𝑡 squared minus 23 𝑡 plus 14 equals zero. Okay, great, so now what we’re gonna do is actually we’re gonna solve this by factoring.

When we factor our equation, what we actually get is eight 𝑡 minus seven multiplied by 𝑡 minus two. And this is all equal to zero. What we’ll do now is going to show you how we got to those factors. So this is factoring with coefficients of 𝑥 squared greater than one. We’ll use a real method for this.

So the first thing we do is we actually multiply. And in this case, it’s the coefficient of 𝑡 squared cause we’re using 𝑡 in our equation. So that’s eight. And then we multiply that by 14. So it’s going to give us 112. So now what we do is we actually find two numbers whose product is 112 and whose sum is the coefficient of 𝑡 which is negative 23. The most two numbers are negative 16 and negative seven. And that’s because negative 16 multiplied by negative seven gives us 112. And we got two negatives. Obviously, we want a positive. And then negative 16 plus negative seven is like negative 16 minus seven is gonna give us negative 23.

Okay, great, so now what we do is we actually substitute negative 16 and negative seven in for our coefficient of 𝑡. So what we actually do is we split up negative 23 into two parts. So we now have the expression eight 𝑡 squared minus 16 𝑡 minus seven 𝑡 plus 14. So now what we’re going to do is we gonna factor each part separately. So we’ll factor the first two terms and factor the second two terms.

So if I factor the first two terms, I’m gonna have eight 𝑡 outside the parentheses. And that’s because there’s eight in both eight and 16 and 𝑡 in both terms. And then inside the parentheses, we’re gonna have 𝑡 minus two.

And then for our second two terms, we’re gonna have negative seven outside the parentheses and that’s because there’s negative seven in both negative seven 𝑡 and 14. And then inside, what we’re gonna have is 𝑡 minus two. And what we should know to this point is that it should be the same as the previous parentheses. And that’s because actually that’s going to be one of our factors.

Great, so now we’ve got this, we can find our factors. So our first parenthesis is eight 𝑡 minus seven. Our second parenthesis is gonna be 𝑡 minus two. Okay, great, so we’ve now shown you how we actually did the factoring, we can actually get on and finish solving the problem.

So now we’ve actually got our equation factored, we can therefore say that 𝑡 is equal to seven over eight and two. So therefore, what we can say is that between the times seven over eight seconds and two seconds, our ball was actually 33 feet above the ground.

So therefore, we can say that the time the ball was 33 feet above the ground is equal to two minus seven over eight because that was the two times that we achieved from solving the equation, which is equal to one and one-eighth seconds.

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