2006-12-14 19:51:52 · 10 answers · asked by Jennifer N 1 in Science & Mathematics ➔ Mathematics
how did you get to
(5v-3)(5v-3)=0
2006-12-14 19:58:14 · update #1
ohhhh ok, so its a little guess and checking?
2006-12-14 20:02:21 · update #2
what is the difference between this question and:
3u^2+4u-15=0 , why doesn't the AC method work with the first question like it does with this question
2006-12-14 20:08:05 · update #3
25v^2 - 30v = -9
First thing you have to do is move the -9 over to the left hand side, forming a quadratic.
25v^2 - 30v + 9 = 0
Now, factor. Note that 25v^2 can be split into 5v and 5v.
(5v + ?) (5v + ?) = 0
Now it's a matter of solving for the question marks. Since the middle term (coefficient of v) is negative, they're likely going to both be negative values. Here's where you ask yourself: what two numbers can you multiply to get 9? The answer to that is -3 and -3. Now, we test the product of the outer values added to the product of the inner values. 5v(-3) + 5v(-3) = -30v, which is our middle term, so that is correct.
(5v - 3) (5v - 3) = 0
To which we equate each to 0, but they are the same factor, so we only use one.
5v - 3 = 0
5v = 3
v = 3/5
2006-12-14 19:58:24 · answer #1 · answered by Puggy 7 · 2⤊ 0⤋
Add 9 to both sides:
25v^2-30v+9=0
Since -3 and -3 equal 9, here's your answer:
(5v-3)(5v-3)
v=3/5
2006-12-15 00:54:50 · answer #2 · answered by Anonymous · 1⤊ 0⤋
The solution for a quadratic equation of the form ax^2+bx+c=0 is:
x=(-b +/- SQR(b^2-4ac))/2a
For the equation 25v^2-30v = -9
we can rewrite
25v^2 - 30v + 9 = 0
then
x= v
a= 25
b= -30
c= 9
The quadratic solution formula then becomes:
v = (- (-30) +/- SQR((-30)^2 - 4(25)(9))) / 2(25)
= (30 +/- SQR(900 - 900)) / 50
v = 0.6
We can test this by putting it back in your original equation:
25 (0.6)^2 - 30(0.6) =
9 - 18 =
-9
So it's the right solution. Obviously, if you're familiar with the factoring, you can see that this is an easy set of factors and do it as the others have done it above.
2006-12-14 20:07:13 · answer #3 · answered by TimmyD 3 · 2⤊ 0⤋
25v^2 - 30v + 9 =0
(5v - 3)^2 = 0
5v -3 = 0
v = 3/5
2006-12-14 20:01:49 · answer #4 · answered by Srinivas c 2 · 2⤊ 0⤋
25v² - 30v = - 9
25v² - 30v + 9 = - 9 + 9
25v - 30v + 9
(5v - 3)(5v - 30)
- - - - - - -
Roots
5v - 3 = 0
5v - 3 + 3 = 0 + 3
5v = 3
5v/5 = 3/5
v = 3/5
- - - - - - -s-
2006-12-14 21:19:22 · answer #5 · answered by SAMUEL D 7 · 2⤊ 0⤋
25v^2 - 30v = - 9
25v^2 - 30v + 9 = 0
(5x - 3)^2 = 0
5x = 3
x = 3/5
2006-12-14 20:00:56 · answer #6 · answered by Helmut 7 · 2⤊ 0⤋
25v^2-30v+9=0
(5v)^2-2(5v)(3)+3^2=0
this is like the identity (a-b)^2=a^2-2ab+b^2
so this is (5v-3)^2=0
so repeated roots
v=3/5,3/5
aliter
25v^2-15v-15v+9=0
5v(5v-3)-3(5v-3)=0
=(5v-3)(5v-3)=0
5v-3=0 v=3/5
5v-3=0 v=3/5
2006-12-14 19:59:40 · answer #7 · answered by raj 7 · 2⤊ 0⤋
get the square root of (25v^2) then subtract 30v you'll have a result of -25v which is equated to 9. dividing the both terms by -25, you'll end up with the answer -0.36. am i correct? can i get a booyah!!??
2006-12-14 20:26:18 · answer #8 · answered by Maria Delia N 1 · 2⤊ 0⤋
x² - 2x -13 = 0 because you could't ingredient this, you want to apply the quadratic formula: Given: ax² + bx + c = 0 x = [-b ± ?(b² - 4ac)] / 2a x² - 2x -13 = 0 a = a million, b = -2, c = -13 x = [-b ± ?(b² - 4ac)] / 2a x = [-(-2) ± ?((-2)² - 4(a million)(-13))] / 2(a million) x = [2 ± ?(4 + fifty 2)] / 2 x = [2 ± ?fifty six] / 2 x = [2 ± ?(4*14)] / 2 x = [2 ± 2?14] / 2 x = a million ± ?14
2016-10-18 07:52:58 · answer #9 · answered by ? 4 · 0⤊ 0⤋
25v^2-30v=-9
25v^2-30v+9=0
(5v-3)(5v-3)=0
5v-3=0
5v=3
v=3/5
2006-12-14 19:55:39 · answer #10 · answered by tma 6 · 3⤊ 0⤋