SOLUTION: Please help. 1. If x = 1 and x = -8, then form a quadratic equation. I have no idea where to even begin an equation. 2. What type of solution do you get for quadratic equations

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help. 1. If x = 1 and x = -8, then form a quadratic equation. I have no idea where to even begin an equation. 2. What type of solution do you get for quadratic equations       Log On


   

Question 149605: Please help.
1. If x = 1 and x = -8, then form a quadratic equation. I have no idea where to even begin an equation.
2. What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.
Iknow if discriminant is less than zero no solutions are defined. But I don't know how to explain why or how to begin an example. Please explain and keep it as simplistic as possible so that I can understand. Like if you were explaining it to a 9 year old.
Thank you so much for all your time and help!!!!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1

Start with the given zeros

x=1 and x=-8


Get all terms to the left side in each case (ie subtract 1 from both sides in the first equation and add 8 to both sides in the second equation)


x-1=0 and x%2B8=0


Now use the zero product property in reverse to join the factors.


%28x-1%29%28x%2B8%29=0


FOIL and multiply


x%5E2%2B7x-8=0


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Answer:

So the polynomial with zeros of x=1 and x=-8 is

x%5E2%2B7%2Ax-8



Notice how if we graph y=x%5E2%2B7%2Ax-8, we can see that the polynomial has roots of x=1 and x=-8

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2B7%2Ax-8+%29+ Graph of y=x%5E2%2B7%2Ax-8 with roots of x=1 and x=-8





# 2

Remember the quadratic formula is x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29


Where the discriminant is D=b%5E2-4ac. So the quadratic formula could also look like x+=+%28-b+%2B-+sqrt%28+D+%29%29%2F%282a%29


If D<0, then we'll be taking the square root of a negative number (which we cannot do). So this results in 2 complex solutions (ie no real solutions).



For example, let's find the discriminant for y=x%5E2%2B2x%2B5


From x%5E2%2B2x%2B5 we can see that a=1, b=2, and c=5


D=b%5E2-4ac Start with the discriminant formula


D=%282%29%5E2-4%281%29%285%29 Plug in a=1, b=2, and c=5


D=4-4%281%29%285%29 Square 2 to get 4


D=4-20 Multiply 4%281%29%285%29 to get %284%29%285%29=20


D=-16 Subtract 20 from 4 to get -16


Since the discriminant is less than zero, this means that there are two complex solutions (or no real solutions)


Now let's use the quadratic formula to find the solutions of y=x%5E2%2B2x%2B5


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%282%29+%2B-+sqrt%28+%282%29%5E2-4%281%29%285%29+%29%29%2F%282%281%29%29 Plug in a=1, b=2, and c=5


x+=+%28-2+%2B-+sqrt%28+4-4%281%29%285%29+%29%29%2F%282%281%29%29 Square 2 to get 4.


x+=+%28-2+%2B-+sqrt%28+4-20+%29%29%2F%282%281%29%29 Multiply 4%281%29%285%29 to get 20


x+=+%28-2+%2B-+sqrt%28+-16+%29%29%2F%282%281%29%29 Subtract 20 from 4 to get -16


x+=+%28-2+%2B-+sqrt%28+-16+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-2+%2B-+4%2Ai%29%2F%282%29 Take the square root of -16 to get 4%2Ai.


x+=+%28-2+%2B+4%2Ai%29%2F%282%29 or x+=+%28-2+-+4%2Ai%29%2F%282%29 Break up the expression.


x+=+%28-2%29%2F%282%29+%2B+%284%2Ai%29%2F%282%29 or x+=++%28-2%29%2F%282%29+-+%284%2Ai%29%2F%282%29 Break up the fraction for each case.


x+=+-1%2B2%2Ai or x+=++-1-2%2Ai Reduce.


So our answers are x+=+-1%2B2%2Ai or x+=+-1-2%2Ai


Since our answers are complex (non real), this verifies our original claim.