Quadratics (1 Viewer)

[Fortian09]

Heya all, more problems...

First one

Find the real root (s) of each of the following equations. (leave answers in 3d.p. if necessay).

1. (1/x) + (1/(x+2) = 2

2. 3^{x^2} = 27

Others to solve:

1. (log x)² = log x²

2. 1/(x-3) - 1/(x-4) = 1/ (x-5)

3. 2^x + 2<sup>2-x] = 5

4. 1/(root((x^2) - 2)) + 1/(root((x^2) + 2))= -3</sup>

 

[Pwnage101]

1). now we need to manipulate this:

(1/x)+1/(x+2)=2 -----> X CANNOT = 0 OR -2 (NOT IN DOMAIN)

(x+2)+x=2x(x+2)

2x+2=2x^2 + 4x

2x^2+2x-2=0

x^2+x-1=0

NB/// (1/x)+1/(x+2)=2 IS NOT THE SAME AS x^2+x-1=0 HOWEVER, THEY HAVE THE SAME ROOTS:

use quadratic formula, and u should get:

x= (-1+root5)/2 OR = (-1-root5)/2


2.

3^(x^2) = 27
take logs of both sides

log(3^(x^2)) = log(3^3), as 27 = 3^3

x^2*(log3) = 3*(log3) by log laws

x^2=3

ie x= root3 OR x=-root3

others:

1. taking 'log' to mean 'log base 10' - NOT the natural logarithm:

(logx)(logx)=log(x^2)
(logx)(logx)=2(logx) by log laws
(logx)(logx)-2(logx)=0
(logx)[(logx)-2]=0

thus, logx = 0 OR logx=2
x=10^0 OR x = 10^2

ie: x=1 OR x=100

2. AS with before, we need to make an eq2uationw ith the same roots [will not be an equivalent equation]

thus 1/(x-3)-1/(x-4)=1/(x-5) ------------>X CANNOT = 3,4 OR 5

multiply everything by (x-3)(x-4)(x-5) and rearrange to get:

x^2-6x+7=0

uusing quadratic formula:

x= (3+root2) OR x=3-root2

3. Not sure how to handle this, at first sight, but upon looking at the graph, it is clear the result is x=0 or x=2

4. Im pretty sure that has no solutions - the left hand side is always positive, as 1/(root of anything) is positive, yet the RHS is negative, thus it will never hold - sketching the graph also yields there ARE NO REAL ROOTS

Hope that helped, and my steps can be easily followed

 

[vafa]

View attachment 16918

 

[lolokay]

2. 3^x2 = 3³
x = +- sqrt3

'Others to solve'
1. (ln x)² = 2ln x
(lnx)² - 2ln x = 0
lnx(lnx -2) = 0
lnx = 0; x = 1
or lnx = 2; x = e²

2. (x-4)(x-5) - (x-3)(x-5) - (x-3)(x-4) = 0
-x² + 6x - 7 = 0
x² - 6x + 7 = 0
x = 6+-sqrt[8]/2 = 3+-sqrt2

3. 2^x + 2^2-x = 5
x = 0, 2 by inspection

2^x + 4*2^-x - 5 = 0
2^2x - 5*2^x + 4 = 0
2^x = 5+-sqrt[9]/2 = 1, 4
x = 0, 2

4. don't the roots imply that LHS is positive, but RHS is negative?

 

[bored of sc]

View attachment 16918

Question 2 is wrong.

It should be:

3^{x^2} = 27
3^{x^2} = 3³
Equate powers:
x² = 3
x = + root3

 

[vafa]

Question 2 is wrong.

It should be:

3^{x^2} = 27
3^{x^2} = 3³
Equate powers:
x² = 3
x = + root3

Thanks for that, I am used to do my mathematics on my Linux machine and I sometimes do such mistakes.

 

[Slidey]

Linux is excellent.