A maths sec 3 replied by SBS261P @ Sun, 20 Jan 2008 10:50:37 +0800

SBS261P — Sun, 20 Jan 2008 10:50:37 +0800

x cannot be 1 becoz

x^1/2 = -1 means

square root x = -1

but square root cannot give negative number so must reject

A maths sec 3 replied by gobez14 @ Sun, 20 Jan 2008 09:28:08 +0800

gobez14 — Sun, 20 Jan 2008 09:28:08 +0800

Originally posted by wishboy:

4x^½ = 3 - 7x

Let x^½ be u

4u = 3 - 7u²
7u² + 4u - 3 = 0
(7u - 3)(u + 1) = 0
7u - 3 = 0 or u + 1 = 0
u = 3/7 or u = -1
x^½ = 3/7 or x^½ = -1
x = 9/49 or 1

FYI, x^½ is equals to "square root of x", written as "sqrt x"

but hor y i sub 1 into the original equation 1 side is 4 the other side is -4 ?
liddat still can??

A maths sec 3 replied by wishboy @ Sat, 19 Jan 2008 20:02:39 +0800

wishboy — Sat, 19 Jan 2008 20:02:39 +0800

Originally posted by gobez14:

can elaborate?

4x^½ = 3 - 7x

Let x^½ be u

4u = 3 - 7u²
7u² + 4u - 3 = 0
(7u - 3)(u + 1) = 0
7u - 3 = 0 or u + 1 = 0
u = 3/7 or u = -1
x^½ = 3/7 or x^½ = -1
x = 9/49 or 1

FYI, x^½ is equals to "square root of x", written as "sqrt x"

A maths sec 3 replied by gobez14 @ Sat, 19 Jan 2008 15:45:10 +0800

gobez14 — Sat, 19 Jan 2008 15:45:10 +0800

Originally posted by SBS261P:

gt careless mistake... shld be

z = 3/7

x = (3/7)^2 = 9/49

can elaborate?

A maths sec 3 replied by ^tamago^ @ Sat, 19 Jan 2008 15:32:22 +0800

^tamago^ — Sat, 19 Jan 2008 15:32:22 +0800

Originally posted by SBS261P:

gt careless mistake... shld be

z = 3/7

x = (3/7)^2 = 9/49

oh yeah. fug~!

A maths sec 3 replied by ^tamago^ @ Sat, 19 Jan 2008 15:31:36 +0800

^tamago^ — Sat, 19 Jan 2008 15:31:36 +0800

Originally posted by gobez14:

how come liddat ar? 7z²+4z-3=0

sqrt(x) = z
x = z²

4*sqrt(x)=3-7x
4z=3-7z²
7z²+4z-3=0
(7z-3)(z+1)=0
z = 3/7 or -1

A maths sec 3 replied by SBS261P @ Sat, 19 Jan 2008 15:30:29 +0800

SBS261P — Sat, 19 Jan 2008 15:30:29 +0800

gt careless mistake... shld be

z = 3/7

x = (3/7)^2 = 9/49

A maths sec 3 replied by gobez14 @ Sat, 19 Jan 2008 15:23:38 +0800

gobez14 — Sat, 19 Jan 2008 15:23:38 +0800

Originally posted by ^tamago^:

3 log(a,2) - 4 + log(a,a³)
= log(a,8) - log(a,a^4) + log(a,a³)
= log[a,(8/a)]

4*sqrt(x)=3-7x
Let z be sqrt(x), z>=0
7z²+4z-3=0
(7z-3)(z+1)=0
z = 3/7 or -1 (rej. as z>=0)
x = sqrt(3/7)

how come liddat ar? 7z²+4z-3=0

A maths sec 3 replied by ^tamago^ @ Sat, 19 Jan 2008 15:04:01 +0800

^tamago^ — Sat, 19 Jan 2008 15:04:01 +0800

3 log(a,2) - 4 + log(a,a³)
= log(a,8) - log(a,a^4) + log(a,a³)
= log[a,(8/a)]

4*sqrt(x)=3-7x
Let z be sqrt(x), z>=0
7z²+4z-3=0
(7z-3)(z+1)=0
z = 3/7 or -1 (rej. as z>=0)
x = sqrt(3/7)

A maths sec 3 replied by gobez14 @ Sat, 19 Jan 2008 14:44:57 +0800

gobez14 — Sat, 19 Jan 2008 14:44:57 +0800

hi

can any1 tell mi wads teh answer of 3 log a (2) - 4 + log a (a to the power of 3) whe nconverted into a single logarithmic form

and value of x when 4 (x to the power of half) = 3-7x

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