SOLUTION: determine all possible values of k so that the quadratic below has one zero.
x^2 + (k-5)x + (4-5k) = 0
the asnwers are k = -1 and k = -9. thank you!
Question 1106233: determine all possible values of k so that the quadratic below has one zero.
x^2 + (k-5)x + (4-5k) = 0
the asnwers are k = -1 and k = -9. thank you! Found 2 solutions by Boreal, ikleyn:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! This needs completing the square, since that would give one zero with multiplicity 2. This is done by taking the coefficient of the x term, dividing it by 2 and squaring it. That will give the coefficient of the constant.
Here, those values are:
[(1/2)(k-5)]^2=(4-5k)
(1/4)(k^2-10k+25)=(4-5k)
k^2-10k+25=16-20k, multiplying by 4
k^2+10k+9=0
(k+9)(k+1)=0
k=-9 and -1
A quadratic equation ax^2 + bx + c = 0
- has two real zeroes if and only if the discriminant d = b^2 - 4ac is POSITIVE;
- has one real zero if and only if the discriminant d = b^2 - 4ac is ZERO;
- has NO real zeroes if and only if the discriminant d = b^2 - 4ac is NEGATIVE.
So, for your equation, in order for the condition was satisfied, the discriminant
(k-5)^2 - 4*(4-5k) must be equal to zero:
(k-5)^2 - 4*(4-5k) = 0.
It is your equation to find the values of "k".
Simplify and solve it by any method you know.
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If you still have questions, you can post them to me through the "Thank you" window/form.
In his case refer to the problem ID number 1106233.
