Studying for Finals and want to understand these. Thank you
2007-05-17 15:37:38 · 13 answers · asked by whattheheck 4 in Science & Mathematics ➔ Mathematics
a² = 5a
a² - 5a = 0
a(a-5) = 0
This can only be true when a=0 or when a=5
2007-05-17 15:41:32 · answer #1 · answered by Astral Walker 7 · 1⤊ 0⤋
a^2 = 5a
a^2 - 5a=0
a(a-5) =0
a=0 or a-5=0
if a-5=0
a= 5
a= 0, 5
2007-05-17 15:43:57 · answer #2 · answered by bob 2 · 0⤊ 0⤋
a^2=5a
=> a^2-5a=0
=>a(a-5)=0
=> a=0 or a=5 are the solutions of the above equation
2007-05-17 15:42:00 · answer #3 · answered by Pavan 3 · 1⤊ 0⤋
ok, i don't know why you trying to square root the problem, i just makes it harder... t^2+6t-16=0 here you have to find 2 numbers that multiplied give 16 and added give 6 and wright it like this, (t-2)(t+8)=0 once its simplified, if you want to solve for t, then split the equation in two: t-2=0 and t+8=0 solve for each: t=2 and t=-8 and those are your answers the other one is done the same way ps: the square root of 16 is 4
2016-05-22 01:46:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋
a^2 - 5 a = 0
a (a - 5) = 0
a = 0 or a = 5
2007-05-17 15:41:55 · answer #5 · answered by Daniel T 2 · 1⤊ 0⤋
Get a^2 and 5a on the same side, and then factor out the common factor a.
2007-05-17 15:41:44 · answer #6 · answered by donaldgirod 2 · 1⤊ 0⤋
a^2=5a
a^2-5a=0
a(a-5)=0
a=0
or
a-5=0 ==>a=5
2007-05-17 21:15:09 · answer #7 · answered by shiva 3 · 0⤊ 0⤋
a^2 - 5a = 0
a(a-5)=0
2007-05-17 15:40:13 · answer #8 · answered by Anonymous · 1⤊ 0⤋
a² - 5a = 0
a.(a - 5) = 0
a = 0, a = 5
2007-05-17 20:10:01 · answer #9 · answered by Como 7 · 0⤊ 0⤋
a^2 - 5a = 0
a(a-5)=0
a=0, a=5
2007-05-17 15:40:58 · answer #10 · answered by fcas80 7 · 1⤊ 0⤋