How do I find the value of K that will make x^2-6x+K a perfect square?
2006-12-01 11:00:17 · 5 answers · asked by c_eva02 2 in Science & Mathematics ➔ Mathematics
ok what about if I have x^2+2kx+75 and i need to find all real values. Do i use the same formula?
2006-12-01 11:15:09 · update #1
x² + bx + c will be a perfect square if c = (b/2)²
so x² - 6x + 9 is a perfect square since 9 = (-6/2)²
2006-12-01 11:11:21 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
x^2-6x+k is a equation of second degre:
to resolve a such equation,we have to set:
b^2-4ac=0 with a=1, b=-6 and c=k
x^2-6x+k is a perfect square if:
b^2-4ac=0
(-6)^2-4(1)(k)=0
36-4k=0 -4k=-36
k=36/4 Then, k=9
x^2-6x+9=(x-3)(x-3)=0
for x=3
3^2-6(3)+9=0
0=0
x^2-6x+k is a perfect square for K=9
2006-12-01 19:35:52 · answer #2 · answered by Johnny 2 · 0⤊ 0⤋
To complete the square, take the coefficient on the x term (-6), divide it in half (-3) and square it (9). That's the number you need to add.
x² - 6x + 9 = (x - 3)²
2006-12-01 19:06:36 · answer #3 · answered by Puzzling 7 · 0⤊ 0⤋
Remember (x + a)^2 = x^2 + 2ax + a^2.
So a=-3 (coefficient of 2nd term) and k=a^2=9.
2006-12-01 19:06:39 · answer #4 · answered by Anonymous · 0⤊ 0⤋
( x - a ) ^2 = x^2 - 2 a x + a^2
For your problem 2ax = 6x => a = 3
a^2 = k = 9
2006-12-01 19:03:34 · answer #5 · answered by Modus Operandi 6 · 0⤊ 0⤋