Given equation (2k+1)x^2 - 3kx + 2k = 4 has different roots, find the range of k. Please help mee
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We can use the fact that if a quadratic equation (ax2+bx+c = 0) has no (real) roots, then the discriminant (b2 - 4ac) should be negative.

In this problem:

a = 2k+1

b = -3k

c = 2k-4

discriminant = (-3k)2 - 4(2k+1)(2k-4)

Simplify.

discriminant = -7k2 + 24k + 16

To determine the intervals, we need to figure out the roots of THIS new quadratic.

k1 = [-24 + sqrt(242 - 4(-7)(16))] / [-14] =  -4/7

k2 = [-24 - sqrt(242 - 4(-7)(16))] / [-14] = 4

So, we've determined the edges of the intervals... And we know if we graphed y = -7x2 + 24x + 16, we would have a downwards facing parabola, because the leading coefficient is negative. So,

-7k2 + 24k + 16 < 0

when

-infinity < k < -4/7  or  4 < k < infinity

Which, in interval notation gives:

(-inf , -4/7) U (4 , inf)

the answer is 489

step-by-step explanation:

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90