I no that for real root
b²≥4ac
If
B²=4ac for equal root
Compare
x²-2x+a=ax²+bx+c
a=1 b=-2 c=a
(2)²=4(a)(a)²
4=4a²
a²=1
a=1
See (1)²-2(1)+1=2-2=0
For which values of a do y = x^2 − 2x + a have zero places?
(Sorry if this is translated wrong, it's nollställen in Swedish, which could be zeros.. Quadratic Function maybe?)
3 answers
Correction
2²=4(a)(a)
4=4a²
Not
2²=4(a)(a)²
2²=4(a)(a)
4=4a²
Not
2²=4(a)(a)²
For real zeros,
b^2 - 4ac ≥ 0
4 - 4(1)(a) ≥ 0
1 - a ≥ 0
-a ≥ -1
a ≤ 1
The confusion arises with the use of "a" , which is used in the definition of
b^2 - 4ac for ax^2 + bx + c = 0
A better wording might have been:
For which values of k do y = x^2 − 2x + k have zero places?
notice in the original we would have
a = 1
b = -2
c = a
b^2 - 4ac ≥ 0
4 - 4(1)(a) ≥ 0
1 - a ≥ 0
-a ≥ -1
a ≤ 1
The confusion arises with the use of "a" , which is used in the definition of
b^2 - 4ac for ax^2 + bx + c = 0
A better wording might have been:
For which values of k do y = x^2 − 2x + k have zero places?
notice in the original we would have
a = 1
b = -2
c = a