1. ## Real-word application of Quadratic Functions & Applying The Square

I am confused with the following question because I think I have to use fractions and I'm having difficulty factoring the fractions here, especially when I have to factor
the trinomial of -5(t2 - 7t + 49/4). And plus.. Can you actually factor time in this real-word application???!! I can't tell if it's a trick question and I'm supposed to use decimals instead of fractions..
By the way, I don't even know if this is Algebra or what is what.. So I apologize if its in the wrong section.

An emergency flare is fired from a vehicle. Its height, h metres, is given by h(t)= -5t2 + 35t + 50, where t is the time in seconds.

a) What is the maximum height reached by the flare? How long does it take to reach the maximum height?

Since h(t)= -5t2 + 35t + 50 is a quadratic function with a=-5, the corresponding graph is a parabola that is concave down. The maximum occurs at the vertex. So to find the maximum height of the rocket, convert the given equation into vertex form.

 h(t) = -5t2 + 35t + 50 1) Common factor -5 from first two terms =-5(t2 - 7t)+ 50 2) Complete the square. b= 1/2 (-7) = -7/ (My question is.. Can't I just use -3.5 instead of using a fraction) b2 = (-7/2)2 = 49/4 = -5(t2 - 7t+ 49/4 - 49/4) + 50 3) Group trinomial = -5[(t2 -7t + 49/4) -49/4] + 50 4) Distribute -5 with brackets and -49/4. -5 x -49/4 = 245/4 =-5(t2 -7t + 49/4) + 245/4 + 50 5) Factor trinomial and find common denominator =-59( t - ? )2 + 245/5 + 200/4

Okay, so this is where I'm confused. How do I factor (t2 -7t + 49/4)... Because -7t has to be a result of 2ab and.. 49/4 has to be a result of b2. (Or am I not even supposed to be using the fraction
in the first place because its not application to the question and it doesn't seem to be factorable? Grrr, so confusing! And for the numbers outside the brackets, would the common denominator result in 245/4 + 200 /4 ???

2. ## Re: Real-word application of Quadratic Functions & Applying The Square

$-5t^2+35t+50=0\\\\-5t^2+35t=-50\\\\t^2+7t=10\\\\t^2+7t+3.5^2=10+3.5^2\\\\(t+3.5 )^2=22.25\\\\ \sqrt{(t+3.5)^2}=\sqrt{22.25}\\\\ t+3.5=\sqrt{22.25}\\\\t=\pm\sqrt{22.25}-3.5$

See if this is enough to clear the problem up for you.

3. ## Re: Real-word application of Quadratic Functions & Applying The Square

Can't edit anymore, but I just noticed that as we divide by -5 on both sides, we of course get -7t not +7t. In the end we get t-3.5=sqrt(22.25) leaving us with +/-sqrt(22.25)+3.5 instead of -

4. ## Re: Real-word application of Quadratic Functions & Applying The Square

Thanks for the reply, I understand what you're saying but I have to solve the equation by "Applying the Square" and changing into vertex form without moving things to the opposite side of the equal sign and without division. I have to solve the equation by finding b2, completing the square, grouping the trinomial, multiplying to remove brackets, and then factoring.. So basically the long way as thats what the lesson is about. :/