4x²-25=0
also, 3a² - a - 4 = 0
(setting y to zero and solving for x)
how do you solve for the roots:?
2009-02-21 4:35 pm
回答 (8)
2009-02-21 4:39 pm
✔ 最佳答案
(2x - 5)(2x + 5) = 0x = +/- 5/2
(3a - 4)(a+ 1) = 0
a = 4/3, -1
2009-02-22 12:40 am
1)
4x^2 - 25 = 0
(2x)^2 - 5^2 = 0
(2x + 5)(2x - 5) = 0
2x + 5 = 0
2x = -5
x = -5/2 (-2.5)
2x - 5 = 0
2x = 5
x = 5/2 (2.5)
= = = = = = = =
2)
3a^2 - a - 4 = 0
3a^2 + 3a - 4a - 4 = 0
(3a^2 + 3a) - (4a + 4) = 0
3a(a + 1) - 4(a + 1) = 0
(a + 1)(3a - 4) = 0
a + 1 = 0
a = -1
3a - 4 = 0
3a = 4
a = 4/3
â´ a = -1 , 4/3
4x^2 - 25 = 0
(2x)^2 - 5^2 = 0
(2x + 5)(2x - 5) = 0
2x + 5 = 0
2x = -5
x = -5/2 (-2.5)
2x - 5 = 0
2x = 5
x = 5/2 (2.5)
= = = = = = = =
2)
3a^2 - a - 4 = 0
3a^2 + 3a - 4a - 4 = 0
(3a^2 + 3a) - (4a + 4) = 0
3a(a + 1) - 4(a + 1) = 0
(a + 1)(3a - 4) = 0
a + 1 = 0
a = -1
3a - 4 = 0
3a = 4
a = 4/3
â´ a = -1 , 4/3
2009-02-22 12:40 am
(2x-5)(2x+5)=0x = +/- 5/2
(3a-4)(a+1)=0
a = -1, 4/3
(3a-4)(a+1)=0
a = -1, 4/3
2009-02-22 12:39 am
4x^2 - 25 = 0
(2x - 5)(2x + 5) = 0
x = 5/2
x = -5/2
----------------------------------
3a^2 - a - 4 = 0
(3a - 4)(a + 1) = 0
a = 4/3
a = -1
(2x - 5)(2x + 5) = 0
x = 5/2
x = -5/2
----------------------------------
3a^2 - a - 4 = 0
(3a - 4)(a + 1) = 0
a = 4/3
a = -1
2009-02-22 12:44 am
The first one is a difference of squares so:
4x^2 - 25 = ( 2x - 5 ) ( 2x + 5 ) = 0
then You can set 2x-5 = 0 and 2x + 5 = 0 to get the two values of x.
The second equation can be also factored to (3a - 4)(a + 1)
and using the same thing get the roots of the second equation.
the first one can be 2.5 or -2.5
the second one can be -1 or 1 1/3
4x^2 - 25 = ( 2x - 5 ) ( 2x + 5 ) = 0
then You can set 2x-5 = 0 and 2x + 5 = 0 to get the two values of x.
The second equation can be also factored to (3a - 4)(a + 1)
and using the same thing get the roots of the second equation.
the first one can be 2.5 or -2.5
the second one can be -1 or 1 1/3
參考: me
2009-02-22 12:43 am
4x^2 - 25 = 0
4x^2 = 25
x^2 = 25/4
x = ± 5/2 {square root both sides}
3a^2 - 1a - 4 = 0
ac = 3(-4) = -12
Factors of -12 that add to -1 are +3 & -4
3a^2 + 3a - 4a - 4 = 0 {rewrite for grouping}
3a(a + 1) - 4(a + 1) = 0
(3a - 4)(a + 1) = 0
a = 4/3, -1
4x^2 = 25
x^2 = 25/4
x = ± 5/2 {square root both sides}
3a^2 - 1a - 4 = 0
ac = 3(-4) = -12
Factors of -12 that add to -1 are +3 & -4
3a^2 + 3a - 4a - 4 = 0 {rewrite for grouping}
3a(a + 1) - 4(a + 1) = 0
(3a - 4)(a + 1) = 0
a = 4/3, -1
2009-02-22 12:43 am
WOW! All the Top Contributors of Mathematics have answered! All of them are correct!!!
2009-02-22 12:42 am
4x^2-25=0 is the difference of squares, so
(2x+5)(2x-5)=0 set each of the factors equal to zero and you get
x= -5/2 and x= 5/2
3a^2 - a - 4 = 0 you just factor the polynomial
(3a-4)(a+1) = 0 and do the same as above, set each of the factors equal to zero, so
a = 4/3 and a = -1
(2x+5)(2x-5)=0 set each of the factors equal to zero and you get
x= -5/2 and x= 5/2
3a^2 - a - 4 = 0 you just factor the polynomial
(3a-4)(a+1) = 0 and do the same as above, set each of the factors equal to zero, so
a = 4/3 and a = -1
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