function f(x)=kx^2+8x+5, what value(s) of k will have A) two distinct solutions ? B) one solution ? C) no solutions ?

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Since our function kx^2+8x+5 is a quadratic equation, the best approach to solving this problem is to use the discriminant to determine the conditions.

Recall that from the quadratic equation, we obtain the discriminant, b^2-4ac, where a, b, and c are directly related to the values of the quadratic function...

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Since our function kx^2+8x+5 is a quadratic equation, the best approach to solving this problem is to use the discriminant to determine the conditions.

Recall that from the quadratic equation, we obtain the discriminant, b^2-4ac, where a, b, and c are directly related to the values of the quadratic function ax^2+bx+c.  In our situation, we have a = k, b = 8, and c = 5.

We know that one of three situations can occur:

a) b^2-4ac > 0 <-- two real solutions

b) b^2-4ac = 0 <-- one real, repeated solution

c) b^2-4ac < 0  <-- two complex solutions

 

Now to find our needed conditions, all we need to do is substitute our known values into the discriminant and solve each equality/inequality:

8^2 - 4(k)(5) = 0  -->  Substituting our known values

64 - 20k > 0 -->  Simplifying

64 > 20k --> Adding 20k to both sides

(64/20) > k --> Dividing both sides by 20

k = 16/5.

 

So we know that k = 16/5 is the parameter for the quadratic to have a real, repeated solution.  Working out the same solution for each inequality will show that when k < 16/5, the quadratic has two real solutions, and when k > 16/5, the quadratic has two complex solutions.

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Find values of k so that `kx^2+8x+5` has two distinct solutions, 1 solution, or no solutions.

We are interested in the determinant. If `ax^2+bx+c=0` then `b^2-4ac` is called the determinant. If the determinant is greater than zero there are two distinct solutions, equal to zero there is exactly one solution, and if less than zero there are no real solutions.

In this case we have the determinant as `8^2-4(k)(5)` or 64-20k.

Now 64-20k=0 implies that `k=16/5` .

Thus for `k<16/5` there are exactly two solutions as the determinant is positive.

For `k=16/5` there is exactly one solution.

For `k>16/5` there are no real solutions.

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