Solve the system of equations.
\[y = -0.04x ^{2} + 8.3x \] y = 1.12x + 1.6
My choices: A. no solution B. (0.22, 1.85) C. (179.28, 202.39) D. (0.22, 1.85) and (179.28, 202.39)
well since they are both equal to y you can set the 2 equations equal to one another and then use the quadratic equation to solve for the solutions
How do I do that @captaincook
do you know the quadratic equation?
no
well then you could always plug the x values into the equations and see if the y values equal what the answer says it is.
Ah, I dont know how to...
how do i solve for x
ok one sec
−0.04x^2+8.3x = 1.12x + 1.6 Your goal is to create a quadratic equations which looks like this: 0 = a number x^2 +/- a number x +/- a number without x
one side needs to be zero
i recommend subtracting 1.12x from both sides. What does the equation look like after that?
7,1784 and 1.6 ?
−0.04x^2+7.18x =1.6
do you understand how I got there?
no exactly. im guessing the 7.18 was part of this 7.1784 and 0.04 for orignial equation. but why did you do that... idk why
−0.04x^2 + 8.3x = 1.12x + 1.6 -1.12x -1.12x ------------------------------------ −0.04x^2 +7.18x = 0 + 1.6 OR −0.04x^2 +7.18x = + 1.6 OR −0.04x^2 +7.18x = 1.6
Goal number one is to create an equation that looks like this: http://2.bp.blogspot.com/_YeNFpQ8GgCk/TMVFdA-C9FI/AAAAAAAACns/dqwZHrwBzL0/s400/Slide3a.jpg
in that picture, a = -0.04 and x^2 follows it b = 7.18 and a solitary x follows it You still need to find c c is your constant number, without a variable
to find c: subtract 1.6 from both sides: −0.04x^2 +7.18x = 1.6 - 1.6 = - 1.6 ---------------------------------------- −0.04x^2 +7.18x - 1.6 = 0 now this your quadratic equation in that picture, a = -0.04 and x^2 follows it b = 7.18 and a solitary x follows it c = -1.6 (and no variable)
What about now? understand so far?
yes im kind of getting this for the most part... but if i had to do this by myself i wouldnt know how to start... thanks for doing all this work
SO goal 2 is to solve for x
there are about 5 ways to do this BUT luckily we don't have to...BECAUSE they give us solutions to test out. Just choose a number in your list and well try it out. pick one...
(there are also like 4 ways to do what we did first as well.....lol)
So this numbers you mean (179.28, 202.39)? wow this is some confusing stuff
Oh trust....its crazy pellet for real...lol
what grade or class are you in?
its really funny that I wrote a cuss word and it wrote pellet hahahahah
but to answer your question...yes but only choose one number not both
12th grade hahaha,, ya i was wondering why you wrote pellet
in order for (179.28, 202.39) to be the correct solution, that you must be able to plug 179.28 in for x in both equations and it =0 AND 202.39 into each equation and it equal 0
if one of them, plugged in does not equal 0...well than theres no need to try the other number because that option is wrong.
and its not yoru answer
x = 202.39 y=−0.04x2+8.3x ...lets start with the one that does NOT have a squared x...because if x = 202.39 and x is squared you must find the square root of x and that sucks. y = 1.12x + 1.6 y = 1.12(202.39) + 1.6
y = 226.6768 +1.6
now remember y = 0 so, 0 = 226.6768 +1.6 o = 228.2768 is so false....lol so there for you dont need to plug it into teh other equation and you dont need to check the other number in your answer option....the option is NOT the right one
Wow! so what about 0.22
Wait a minute! if your equation is this: y = 1.12x + 1.6 than that means that 1.12 times anything plus anything has to equal 0 to be the correct answer....the only way that can happen is if x is negative or a 0. NONE of your solutions offer that.
try it with your calculator... i know u have one...lol
yes i have a calculator.
1.12 x 0.22 + 6 = 1.84
So the answer is no solution then
....i think so....my brain is fried...lol
its past midnight here
haha mine too! Thanks for all your help and explaining! I wush I could give ou more than a medal
hang on a sec
stuff that might help you in the class: http://www.youtube.com/watch?v=KNwwu5wjcaA http://www.youtube.com/watch?v=1T-rsltsWnM -----> http://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq
use khan academy for evvveeerrryyyyttthhhiinnggg, lol.....all classes and stuff
Thanks very much
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