How do I work out this problem?
Using the quadratic equation x2 – 4x – 5 = 0, perform the following tasks
b)Solve by using the quadratic formula
showing work.
2007-02-23 10:50:00 · 7 answers · asked by donmigeul 2 in Science & Mathematics ➔ Mathematics
-b +/- square root of (b^2 -4ac)/2a
-(-4) +/- square root (16-20) / 2
4 +/- 1
= -1 or 5
2007-02-23 10:53:18 · answer #1 · answered by csucdartgirl 7 · 1⤊ 0⤋
The quadratic equation is (-b+/-(b^2-4ac))^1/2/2a
so in your equation a=1, b=-4 and c=-5, if you put this values in your equation:
=-(-4) +/- square root((-4)^2-4(1)(-5))
2(1)
=4 +/- square root(16+20)
2
=4 +/- square root(36)
2
=4 +/- 6
2
there are two results
x1=10/2=5
x2=-2/2=-1
(5)^2-4(5)-5 = 25-20-5=0
(-1)^2-4(-1)-5 = 1 + 4 -5 = 0
2007-02-23 11:08:23 · answer #2 · answered by Isa 2 · 0⤊ 0⤋
x=0
2007-02-23 10:58:26 · answer #3 · answered by Mz Toy 2 · 0⤊ 0⤋
X1 = [4 + sqrt(16 +4*1*5) ]/2 = (4 + 6 )/ 2= 5
X2 = [4 - sqrt(16 +4*1*5) ]/2 = (4-6)/2 = -2/3
i think i made an error
2007-02-23 11:00:24 · answer #4 · answered by gjmb1960 7 · 0⤊ 0⤋
Surely you can find the quadratic equation in your book or on the web. If you cannot, then a solution would be worthless. If you can, then substitute a, b, and c.
2007-02-23 10:53:07 · answer #5 · answered by Fred 7 · 0⤊ 0⤋
Well this is correct answer,
the eqn is in the form of ax^2+bx+c=0,
By comparing with the above one we can understand that
a=1
b=-4and
c=-5
and Now,
x = {-b +(-)sqrt(b^2-4ac)}/2a
By substituting the values of a,b,c in the above eq^n,We could get x as
x=5
x=-1
2007-02-23 11:09:31 · answer #6 · answered by MJ 2 · 1⤊ 0⤋
(x - y^5)^3 = x^3 - 3x^2y^5 + 3xy^10 - y^15
2016-05-24 03:42:40 · answer #7 · answered by Anonymous · 0⤊ 0⤋