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2007-01-02 11:35:06 · 8 answers · asked by PORCHA B 1 in Science & Mathematics Mathematics

8 answers

2t^2 + 10t - 48 = 0
(2t - 6)(t + 8) = 0
(2t - 6) = 0 or (t + 8) = 0
2t = 6, so t = 3 or t = -8

t = 3 or -8

2007-01-02 11:43:57 · answer #1 · answered by Tom :: Athier than Thou 6 · 1 0

First: find a number divisible by the coefficients which is 2 > Factor out 2 and multiply it by the remaining terms to get 2t^2 + 10t - 48 >

2(t^2 + 5t - 24) = 0

Sec: factor the terms in the parenthesis:

2(t + 8)(t - 3) = 0

Third: solve for "t" > take both sets of parenthesis and equal them to "0" >

t + 8 = 0
t + 8 - 8 = 0 - 8
t = - 8

t - 3 = 0
t - 3 + 3 = 0 + 3
t = 3

t = -3 and 8

2007-01-02 12:31:25 · answer #2 · answered by ♪♥Annie♥♪ 6 · 0 0

2t^2+10t-48=2(t^2+5t-24)
=2(t+8)(t-3)
=0.

By the zero product property, the product of two real numbers is zero if and only if at least one of the two numbers is zero: ab=0 -> a=0 or b=0.

So,

t+8=0 ->t=-8 or t-3=0 ->t=3.

Thus, t=-8 or t=3.

It is very important to remember that the zero product property is only valid for products involving zero. _Never_ apply it to a product involving a number involving a number other than zero.

Good luck!

2007-01-02 11:44:05 · answer #3 · answered by Anonymous · 1 0

2t^2+10t-48=0 divide by 2
t^2+5t-24=0
(t-3)(t+8)=0
t-3=0
t=3
t+8=0
t=-8

t=3, -8

2007-01-02 11:41:06 · answer #4 · answered by yupchagee 7 · 1 0

2t^2+10t-48=0
2(t^2+5t-24)=0
2[(t-3)(t+8)]=0
So, t-3 or t+8 = 0
t-3 = 0
t=3

t+8 = 0
t=-8

So, t=3 or -8

2007-01-02 11:56:41 · answer #5 · answered by danjlil_43515 4 · 1 0

(2t-6)(t+8)
2(t-3)(t+8)
t=3 or -8

2007-01-02 11:40:01 · answer #6 · answered by raj 7 · 1 0

2t^2+10t-48=0
4t+10t-48=0
14t-48=0
14t=-48

t= -3.43

2007-01-02 11:38:32 · answer #7 · answered by AYBZ 2 · 0 2

(2t+16)(t-3)=0
2t+16=0
2t=-16
t=-8,3

2007-01-02 11:41:44 · answer #8 · answered by Dave aka Spider Monkey 7 · 1 0

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