# solve the equationWatch

This discussion is closed.
#1
square root x - (6/ squarerootx) = 1

?

firstly multiply by sqrt of x giving you

x - 6 = 1rtx

then square or take 6 across?

help?
0
13 years ago
#2
(Original post by thewebmaster)
square root x - (6/ squarerootx) = 1

?

firstly multiply by sqrt of x giving you

x - 6 = 1rtx

then square or take 6 across?

help?
substitute rtx =t

t -(6/t) =1
tÂ² - 6 - t = 0
(t-3)(t+2) =0
t=3
x=9

Aitch

Assumed: you don't want rtx=-2 as a solution...
0
13 years ago
#3
Square without taking 6 across. Check solutions afterwards.
0
13 years ago
#4
Substitute sqrtx = a.
aÂ² - 6 = a
aÂ² - a - 6 = 0
(a+2)(a-3) = 0

a=-2 or a=3
But a = sqrtx. So x = 4 or 9. (x=4 if you take negative square root, so I don't reckon you'd use that at a valid answer..)
0
13 years ago
#5
x - 6 = 1rtx
xÂ² - 12x + 36 = x (square both sides)
xÂ² - 13x + 36 = 0
(x-4)(x-9) = 0
x=4 or x=9.
Again, 4 if you take the negative square root, 9 if the positive.
0
13 years ago
#6
x=9 (see above - hehe I even used 'a' too!), as since it doesn't specifically say -ive square root in the question. I think - feel free to disagree!
0
13 years ago
#7
lol... it's funny when someone posts a question and about 3 people end up typing pretty much the same answer at the same time.
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13 years ago
#8
. It can be really annoying if it's a really long answer and you find somebody did it about 10 minutes faster than you or something!
(the other annoying thing is that you only get rep for about 1/20 of the questions you answer... or at least I do )
0
#9

BRILLIANT
I have a simultaneous equation if anyone fancies a go at helping me here..

2x + y = 1
xÂ² + xy - 4yÂ² = 2

I know its got something to do with letting y = something? perhaps...

y = 1 - 2x

substitute this in....

xÂ² + (x)(1-2x) - 4(1-2x)Â² = 2
Solve for x? substitute back in to original formula for y? (I AM NOT SAYING THIS IS CORRECT! IN FACT AT THE MOMENT ITS JUST FLOWING FROM NOWHERE)

xÂ² + x - 2xÂ² - 4(1 DIFFERENCE OF TWO SQUARES HERE? CANCEL) = 2
xÂ² + x - 2xÂ² - 4 = 2

-xÂ² + x - 4 = 2 (+xÂ²)
normal quadratic? but a is negative therefore quadratic is upside down? so i simply added + xÂ² to both sides, i'm allowed to do this aren't i?

--> 2 + xÂ² - x + 4
--> xÂ² - x + 6

0
13 years ago
#10
(Original post by thewebmaster)

BRILLIANT
I have a simultaneous equation if anyone fancies a go at helping me here..

2x + y = 1
xÂ² + xy - 4yÂ² = 2

I know its got something to do with letting y = something? perhaps...

y = 1 - 2x

substitute this in....

xÂ² + (x)(1-2x) - 4(1-2x)Â² = 2
Solve for x? substitute back in to original formula for y? (I AM NOT SAYING THIS IS CORRECT! IN FACT AT THE MOMENT ITS JUST FLOWING FROM NOWHERE)

xÂ² + x - 2xÂ² - 4(1 DIFFERENCE OF TWO SQUARES HERE? CANCEL) = 2
xÂ² + x - 2xÂ² - 4 = 2

I'm happy up until this point:
xÂ² + (x)(1-2x) - 4(1-2x)Â² = 2

Now carrying on, you should get:

xÂ² + (1-2x)(x - 4(1-2x)) = 2
xÂ² + (1-2x)(9x - 4) = 2
xÂ² + (9x - 4 - 18xÂ² + 8x) = 2
-17xÂ² + 17x - 6 = 0
17xÂ² - 17x + 6 = 0

x = (-1/2) +/- rt(119)/34

y = 1 - 2x, so

y = 2 -/+ rt(119)/17

Galois.
0
#11
maybe this bit got copied down wrong, the answers are whole numbers so the quadratic formula shouldn't be needed.

xÂ² + ((x)(1-2x)) - (4(1-2x)Â²) = 2

so

xÂ² + x - 2xÂ² - 4(1-2x)(1-2x) = 2

xÂ² + x - 2xÂ² - 4(1-2x-2x+4xÂ²) = 2

xÂ² + x - 2xÂ² - 4 + 8x + 8x - 16xÂ² = 2

-5xÂ² + x - 4 + 16x -16xÂ²= 2

-17xÂ² + 17x - 6 = 0

divide by 17?

quadratic formula gives a decimal result i think, also the value of A is still negative?
0
13 years ago
#12
(Original post by thewebmaster)
maybe this bit got copied down wrong, the answers are whole numbers so the quadratic formula shouldn't be needed.

xÂ² + ((x)(1-2x)) - (4(1-2x)Â²) = 2

so

xÂ² + x - 2xÂ² - 4(1-2x)(1-2x) = 2

xÂ² + x - 2xÂ² - 4(1-2x-2x+4xÂ²) = 2

xÂ² + x - 2xÂ² - 4 + 8x + 8x - 4xÂ² = 2

-5xÂ² + x - 4 + 16x = 2

-5xÂ² + 17x - 6 = 0

quadratic formula gives a decimal result i think, also the value of A is still negative?
you havent multiplied the 4x^2 in brackets by 4 when you expanded the brackets.
0
13 years ago
#13
(Original post by thewebmaster)

BRILLIANT
I have a simultaneous equation if anyone fancies a go at helping me here..

2x + y = 1
xÂ² + xy - 4yÂ² = 2

I know its got something to do with letting y = something? perhaps...

y = 1 - 2x

substitute this in....

xÂ² + (x)(1-2x) - 4(1-2x)Â² = 2
Solve for x? substitute back in to original formula for y? (I AM NOT SAYING THIS IS CORRECT! IN FACT AT THE MOMENT ITS JUST FLOWING FROM NOWHERE)

xÂ² + x - 2xÂ² - 4(1 DIFFERENCE OF TWO SQUARES HERE? CANCEL) = 2
xÂ² + x - 2xÂ² - 4 = 2

-xÂ² + x - 4 = 2 (+xÂ²)
normal quadratic? but a is negative therefore quadratic is upside down? so i simply added + xÂ² to both sides, i'm allowed to do this aren't i?

--> 2 + xÂ² - x + 4
--> xÂ² - x + 6

there is no real answer to these simultaneous equations - plot a graph and see. you must have typed it out wrong.
0
#14
Here's the actual equation...

2x + y =1
xÂ²+ xy + 3x - y = 4

ok that's a big one the bottom one but i presume its the same method of getting 1st in terms of x substitution and follow through?

I'll have a go.
0
13 years ago
#15
(Original post by thewebmaster)
Here's the actual equation...

2x + y =1
xÂ²+ xy + 3x - y = 4

ok that's a big one the bottom one but i presume its the same method of getting 1st in terms of x substitution and follow through?

I'll have a go.
Yes exactly. From the 1st equation let y = 1 - 2x, and substitute that into the second question.

Also you seem to dislike the fact that a is negative. Do you mean the constant that's infront of xÂ²? In this case, just multiply by -1 to get the xÂ² positive. It doesn't make a difference at all.

Galois.
0
#16
i got one of the answers!

I got one of the pairs of answers namely i got (x =5 , y = -9)

I am struggling with the other set of coordinates though this is what i did. I've impressed myself here...

ok let y = 1- 2x giving you....

xÂ²+x(1-2x)+3x - (1-2x) = 4
xÂ²+x-2xÂ²+3x - 1 -2x = 4 (take the -1 across)
FACTORISE OUT x (thought I was being fancy here) = 5
x(x+1 - 2x + 3 - 2) =5
x(x+1 - 2x + 1) =5
x(x-2x + 2) = 5
x(-x + 2) = 5

therefore x = 5 substitute into one of original equations y = - 9

OK i've got that easily because its just the x, my question is

(-x+2) = 5 how do i use the bracket to find the other value of x and y. or is this all wrong?
0
#17
#

I GOT IT
i know exsactly where i went wrong, (i think)

what i did was
xÂ² + x - 2xÂ² + 3x -1 -2x = 4

what i should have noticed (silly me)
xÂ² + x(1-2x) + 3x - (1 - 2x)

there is a double negative - (1 -2x)
This makes the 2x positive and if you'd like to follow through with me now, sing along if you want...

xÂ²+ x - 2xÂ² + 3x - 1 + 2x = 4
TAKE 1 across as it just gets in the way

xÂ² + x - 2xÂ² + 3x + 2x = 5
Factorise
x(x + 1 - 2x + 3 + 2) = 5
Sort it out a bit
x(x +5 - 2x) = 5
still sorting it out (hoping this works like i seen it before)
x(-x + 5) = 5
-x = 5 - 5
LOST IT AGAIN GOD MIGHTY ARGHH ANYONE HELP ME FROM HERE, am i any more right?
0
13 years ago
#18
you finish up with x^2 - 6x + 5 = 0

factorising...

(x-1)(x-5) = 0

so x = 1 or 5.

then you plug these values into one of the equations to find y.

your way of expanding and simplifying brackets seems a bit crazy, just expand everything, then group together the different powers of x.
0
13 years ago
#19
make one side equal to 0...
0
13 years ago
#20
Yeah, you can't say x(x +5 - 2x) = 5 therefore x=5 or x+5-2x =5, it only works with zero, because of the special property that anything multiplied by zero equals zero.
0
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