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Mathematics 15 Online
OpenStudy (anonymous):

CAN SOMEONE PLEASE HELP ME ???? THIS IS SO CONFUSING. PLEASE PLEASE PLEASE HELP ME? A quadratic equation is shown below: x^2 + 5x + 4 = 0 Part A: Describe the solution(s) to the equation by just determining radicand. Show your work. Part B: Solve 4x^2 -12x + 5 = 0 using an appropriate method. Show the steps of your work, and explain why you chose the method used. Part C: Solve 2x^2 -10x + 3 = 0 by using a method different from the one you used in Part B. Show the steps of your work.

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

anybody?

OpenStudy (campbell_st):

the radicand is the value inside the square root if you are using the general quadratic formula... its commonly known as the discriminant \[\Delta = b^2 - 4ac\] you have a = 1, b = 5 and c = 4 so substitute them and get a value

OpenStudy (anonymous):

oh ok srry wait a sec

OpenStudy (campbell_st):

the conditions for the discriminant \[\Delta > 0 \] two unequal roots... if it is a perfect square the then the roots are rational. \[\Delta = 0\] two equal or repeated roots... the quadratic factors to a perfect square. \[\Delta < 0\] the roots and complex... and the term no real roots is used to describe them. hope it helps

OpenStudy (anonymous):

5^2 - 4 (1) (4)

OpenStudy (campbell_st):

thats correct... so what value does it have...?

OpenStudy (anonymous):

-6

OpenStudy (campbell_st):

oops should read \[\Delta > 0\] two real unequal....

OpenStudy (campbell_st):

ummm try again 25 - 16 =

OpenStudy (anonymous):

oh srry did it in my head real quick

OpenStudy (anonymous):

9

OpenStudy (campbell_st):

great so 9 > 0 so real unequal roots and 9 is a perfect square so the roots are rational. that's it

OpenStudy (anonymous):

ok so that 's part a right ?

OpenStudy (campbell_st):

that's correct...

OpenStudy (anonymous):

ok could you help me with part b and c if not thats fine

OpenStudy (campbell_st):

ok... so for B.... you could use the general quadratic formula \[x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\] a = 4, b = -12 and c = 5 but it can be factored...

OpenStudy (campbell_st):

it depends on when method you are confident with.

OpenStudy (anonymous):

ok so for part c i could factor it right?

OpenStudy (campbell_st):

I'd factor B and use the General Quadratic formula for C

OpenStudy (anonymous):

ok great thank you uhm do u mind helping me with one more i got most parts down but some i dont get

OpenStudy (campbell_st):

ok... what do you need a hand with..

OpenStudy (anonymous):

A Labrador leaps over a hurdle. The function f(t) represents the height of the Labrador above the ground, in inches, at t seconds: f(t) = -16^t2 + 20t A foxhound jumps over the same hurdle. The table shows the height of the foxhound above the ground g(t), in inches, at t seconds: Time (t) g(t) 0 0 0.4 7.44 0.6 9.24 0.75 9.76 1.0 9 1.50 0 Part A: Compare and interpret the maximum of f(t) and g(t)? (4 points) Part B: Which function has a greater x-intercept? What do the x-intercepts of the graphs of f(t) and g(t) represent? (4 points) Part C: Determine the y-intercepts of both functions, and explain what this means in the context of the problem. (2 points)

OpenStudy (anonymous):

OpenStudy (anonymous):

srry im not typing something is wrong with my internet

OpenStudy (anonymous):

it keeps saying that im typing but im not

OpenStudy (campbell_st):

this looks like a test... the max height for g(t) can be read off the table max height is 9.76 when t = 0.75 for f(t) you need to find the line of symmetry, the max height is on this line \[t = \frac{-b}{2a}\] you have b = 20 and a = -16 so substitute and find the value of t, then substitute it into f(t) to find the max height. of now for part B 0 = -4t(4t - 5) find the values of t than make the equation true... these are the times when the dog and fox are on the ground. In both cases there is a solution t = 0.... you need to find the other value in g(t) find the t values where g(t) = 0 hope it helps

OpenStudy (anonymous):

yeah it is a test hope you dont mind

OpenStudy (campbell_st):

I do.. so I'll leave you to finish it...

OpenStudy (anonymous):

ok thanks tho u have been a great help

TheSmartOne (thesmartone):

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TheSmartOne (thesmartone):

@dfjaijdkjlakjf

OpenStudy (anonymous):

ok im srry @TheSmartOne

OpenStudy (anonymous):

but he explained it to me

TheSmartOne (thesmartone):

Yes but posting test questions is cheating...

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