2 rational roots
2 irrational roots
2 complex (conjugate)
real (rational)
2006-11-20 08:33:28 · 8 answers · asked by Dre 1 in Science & Mathematics ➔ Mathematics
Well, this is a standard square trinomial, which you can tell because it is written in the form a^2x^2 + 2abx + b^2, the square root of which is ax + b. In this case, that means that your new equation is:
(4y - 1)^2 = 0
Take the square root of both sides to get:
4y - 1 = 0
4y = 1
y = 1/4
One rational root
2006-11-20 08:42:06 · answer #1 · answered by bgdddymtty 3 · 1⤊ 0⤋
It has 2 rational, equal roots, because
16y² -8y +1 = (4y -1)². So each root is 1/4.
2006-11-20 10:02:54 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
16y^2-8y+1=0
(4y-1)(4y-1)=0
y=1/4 or 1/4
y=1/4
Therefore the equation as 2 repeated roots of 1/4. 1/4 is also a rational number and is real.
Thus technically has 1 real rational root (which is repeated)
2006-11-20 08:42:11 · answer #3 · answered by Oz 4 · 0⤊ 0⤋
Without putting pencil to paper, I see that (b^2-4ac) = 0, so there will be 2 identical rational roots.
2006-11-20 08:39:08 · answer #4 · answered by Steve 7 · 0⤊ 0⤋
2 real rational roots as b^2=4ac
2006-11-20 08:38:58 · answer #5 · answered by raj 7 · 0⤊ 0⤋
It has one real root (otherwise known as two repeated real roots) which is y=1/4 (therefore also rational).
2006-11-20 08:38:55 · answer #6 · answered by martina_ie 3 · 0⤊ 0⤋
Use this formula:
b^2-4ac
a=16,b=-8,c=1
(-8)^2-4(16*1)
64-64=0
One real number(rational)
2006-11-20 08:39:17 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Well, it breaks down to (4y - 1)^2 = 0, whatever that fits into.
2006-11-20 08:37:24 · answer #8 · answered by Chris J 6 · 0⤊ 0⤋