Quadratic word problem

You are stumbling around because you do not understand the concepts you need to apply.

Basic concept 1. A quadratic may have two real solutions, one real solution, or no real solution. If you think in terms of one real solution, you may miss the correct answer.

Basic concept 2.

[MATH]a * b = 0 \implies a = 0 \ne b \text { or } a \ne 0 = b \text { or } a = 0 = b.[/MATH]
If the product of two numbers is zero, at least one must be zero, but it is not necessary for both to be zero.

You are correct that you solve for zero height. But what is the height of the ball at time 0? Why does that solution make sense? Why does that solution not help you solve the question you were asked?

Anyway you get to [MATH]t(30.6 - 4.9t) = 0.[/MATH]
One way that can be true if for t to equal zero? But we know that solution is not relevant to the question. That's OK. There is another solution.

[MATH]30.6 - 4.9t = 0.[/MATH]
Can you solve that equation?
 
Where did that t2 come from in your second picture?

Frankly, I would give both the exact answer and the approximation.

[MATH]t = \dfrac{30.6}{4.9} \approx 6.2[/MATH]
The reason is that I can put the exact answer back into the original equation and check it.

[MATH]30.6 * \dfrac{30.6}{4.9} - 4.9 * \left ( \dfrac{30.6}{4.9} \right )^2 = \dfrac{30.6^2}{4.9} - \dfrac{\cancel {4.9} * 30.6^2}{4.9^{\cancel 2}} = 0.[/MATH]
 
omgineedhelp123 said:
I’m so confused now, I’m sorry. This is the equation
Click to expand...
[MATH]-4.9t^2 + 30.6t = 0 \implies 30.6t - 4.9t^2 = 0 \implies t(30.6 - 4.9t) = 0.[/MATH]
Any problems with that?

Therefore there are TWO REAL SOLUTIONS.

[MATH]\text {Either } t = 0 \text { or } 30.6 - 4.9t = 0.[/MATH]
Do you understand why? That is half the point of this exercise.

t = 0 is a valid solution of the equation. but not a valid answer to the question. Why?

So the answer is that [MATH]30.6 - 4.9t = 0[/MATH].

Solve that simple linear equation for t.

If you can answer my questions, you should not be confused.
 
Yea I understand now when we factored the ‘t’ out of the equation there was still a part of the equation that needed to be solved which was (30.6 -4.9) after we solved that, I got x1= 0 and x2= 6.2
 
30.6 - 4.9t is what you meant, I hope.

But why do we solve the equation [MATH]30.6 - 4.9t = 0.[/MATH] What is the relevance of that 0?

And why do we reject t = 0 as an answer?
 
omgineedhelp123 said:
Is it because there are 2 real solutions because it’s a quadratic function
Click to expand...
That is not quite right. There are at most 2 real solutions to a quadratic equation. There may be 2, 1, or 0.

Here is why factoring is ONE way to solve a quadratic equation. We have

[MATH]30.6t - 4.9t^2 = 0.[/MATH]
The expression on the left is easy to factor.

[MATH]\therefore t(30.6 - 4.9t) = 0 \text { because } t(30.6 - 4.9t) = 30.6t - 4.9t^2[/MATH]
But, and this is a basic principle, when the product of two numbers is zero, at least one of the numbers must be zero.

That means

[MATH]t = 0 \text { or } 30.6 - 4.9t = 0 \implies t = \dfrac{30.6}{4.9}.[/MATH]
Two solutions. Why is the t = 0 solution not a valid answer to the question? I mean it is obvious that

[MATH]30.6 * 0 - 4.9 * 0^2 = 0 - 0 = 0.[/MATH]
It is clearly a valid solution. Why is it not an answer?
 
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