Find a positive value of k such that the equation x^2+kx+9=0 has a repeated solution.
Not looking for the answer per say but how to do it. Also would I be able to do this problem on a TI-83 Calculator?
2006-09-25 05:16:42 · 4 answers · asked by ? 3 in Education & Reference ➔ Homework Help
No you cant. The values inputed to the calculator for forming the quadratic must be numbers not variables like k
2006-09-25 05:27:43 · answer #1 · answered by Karthik I 2 · 1⤊ 1⤋
To do this, take the square root of the constant (9), which is 3.
Why? "repeated solution" essentially means that you're finding the value of K that results in a perfect square binomial.
Then, double the square root you found above (3 * 2 = 6), because when you square a linear equation (such as x+a), you get the binomial x + 2ax + a^2.
k therefore either equals -6 or 6. Since you're looking for the positive value, the answer is 6.
2006-09-25 12:28:27 · answer #2 · answered by ³√carthagebrujah 6 · 1⤊ 1⤋
To have a repeated solution, b^2-4ac = 0
so b=k, a=1, and c=9,
Substitute and solve:
k^2-4(1)(9) = 0
k^2-36=0
k^2=36
so k= 6 and -6
Both values of k are valid answers. If you have +3 and +3 as your solutions, then they add to +6 and multiply to +9, and if you have -3 and -3 as your solutions, then they add to -6 and multiply to +9.
Hope this makes sense. Good Luck.
2006-09-25 12:38:28 · answer #3 · answered by SmileyGirl 4 · 1⤊ 0⤋
for repeated solution b^2=4ac
k^2=4*1*9
k=+/-6
so for k=+/-6 the equation will have
repeated roots
2006-09-25 12:23:27 · answer #4 · answered by raj 7 · 0⤊ 0⤋