Inter. Algebra: find k so 3x^2+10x+k=0 has 1 rtnl soln

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Helen

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Oct 28, 2007
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I am not really sure of this equation. Could someone help?
Find the value of k so that the equation will have exactly one rational solution.
3x ^ (2) + 10x + k = 0
This is what I did.
3x ^ (2) + 10x + k = 0
3x ^ (2) + 10x = - k add 3 to each side.
x ^ (2) + 1/2(10) = - 5 and - 5 ^ (2) = 25/3
answer, hopefully, is( - 25 / 3)
 
Re: Intermediate Algebra

I think k=+25/3.

I used the quadratic formula. b^2-4ac must = 0.

b=10, a=3, c=k.

Supper's on. Gotta go.
 
Re: Intermediate Algebra

Thank you Loren,
I will try it your way. Happy eating. Helen
 
This is what I did.
3x ^ (2) + 10x + k = 0
3x ^ (2) + 10x = - k add 3 to each side. <<<If you add 3 to each side you get 3x^2 + 10x + 3 = -k + 3
x ^ (2) + 1/2(10) = - 5 and - 5 ^ (2) = 25/3
answer, hopefully, is( - 25 / 3)

Looks like maybe you are trying to complete the square???

3x ^ (2) + 10x + k = 0
Divide through by 3.
x^2 + (10/3)x = -k/3
x^2 + (10/3)x + 25/9 = 25/9 - 3k/9
\(\displaystyle (x+\frac{5}{3})^2 =\frac{25-3k}{9}\)
\(\displaystyle \frac{25-3k}{9}=0\)
25-3k=0
k=25/3
 
Helen, you are really taking the long way around on this problem.

The expression under the radical sign in the quadratic formula, b[sup:2brbsqwf]2[/sup:2brbsqwf] - 4ac, tells you about the number and nature of the solutions to a quadratic equation.

If b[sup:2brbsqwf]2[/sup:2brbsqwf] - 4ac = 0, the quadratic has exactly one rational solution.
If b[sup:2brbsqwf]2[/sup:2brbsqwf] - 4ac > 0, the quadratic has exactly two different real solutions, and further, if b[sup:2brbsqwf]2[/sup:2brbsqwf] - 4ac is a perfect square, those solutions will be rational.
If b[sup:2brbsqwf]2[/sup:2brbsqwf] - 4ac < 0, the quadratic has exactly two complex solutions.

In your equation, a = 3, b = 10, and c = k. You want one rational solution. So,

b[sup:2brbsqwf]2[/sup:2brbsqwf] - 4ac = 0
10[sup:2brbsqwf]2[/sup:2brbsqwf] - 4(3)(k) = 0
100 - 12k = 0

Now....solve that for k.
 
mrspi,
Thank you for your help, I will go ahead and solve it for k. Helen
 
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