SOLUTION: Find the values of k such that 4x^2+16x+k has the following solutions. a. no real solutions

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Question 1179334: Find the values of k such that 4x^2+16x+k has the following solutions.
a. no real solutions
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

the values of k such that 4x%5E2%2B16x%2Bk has the following solutions.
a. no real solution
it will be if discriminant b%5E2-4ac%3C0
4x%5E2%2B16x%2Bk...in this case a=4, b=16, and c=k
16%5E2-4%2A4%2Ak%3C0
16%5E2-16k%3C0
16%2816-k%29%3C0
will be equal to zero if 16-k%3C0->16%3Ck or k%3E16
let try k=17
4x%5E2%2B16x%2B17
and real solution is
4x%5E2%2B16x%2B17=0 use quadratic formula
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29............a=4, b=16, and c=17
x=%28-16%2B-sqrt%2816%5E2-4%2A4%2A17%29%29%2F%282%2A4%29
x=%28-16%2B-sqrt%28-16%29%29%2F8
x=%28-16%2B-4i%29%2F8
x=%28-4%2B-i%29%2F2
x=-2%2Bi%2F2
or
x=-2-i%2F2
see the graph below:

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C4x%5E2%2B16x%2B17%29+