I want to solve a methametical question... a+b=6, 3ab=c find C. Any one can help.(Quardatic formula)?

Question

Infact i know the answer, but to reach the soloution, methematical steps require...I tell you answer is 15

I am explaining the question in words...

There is a two digit number such that if you take sum of these two digits the answer is 6 while the product of these two digits is 1/3 of the original number.

Answer 1

Let your 2-digit number be represented by 10A + B.

Then, we are given :
(1) A + B = 6
and
(2) AB = (10A + B) / 3

From (1) we get B = 6 - A

Substituting this into (2) gives :
A(6 - A) = (10A + 6 - A) / 3

6A - A^2 = (9A + 6) / 3 = 3A + 2

Therefore, A^2 - 3A + 2 = 0
Factorise to give : (A - 1)(A - 2) = 0

Thus A = 1 or 2.

If A = 1, then B = 5.
If A = 2, then B = 4

There are thus 2 solutions for AB.
AB = 15 or 24.

(It helped to see the question in words)

Answer 2

As has been pointed out, you cannot solve this completely
since there are 3 unknowns but only 2 equations. You can, at best however solve for a set of solutions. For example let's
restrict our problem to a and b being whole poisitive numbers.
Then (a,b) could be (1,5) and c is 3(1)(5)=15. A table of solutions follows.
a.. b.. c
0.. 6.. 0
1.. 5.. 15
2 .. 4 .. 24
3.. 3 .. 27
4 .. 2 ... 24
5.. 1 .. 15

Answer 3

from 3ab=c, we get a=bc/3,
substitute this in the other eqn

(bc/3)+b= 6

taking b as common outside,

b(c/3+1)=6

now we have two solutions, either b=6 or c/3+1=6

thus c/3=5 and so

c=15

Answer 4

a + b = 6 then a = 6 - b

3ab = c

3(6-b)b = c substituting for a
18b -3b^2 = c working out the parenthesis
3b^2 - 18b + c = 0 moving it all on one side of the equation

Now you can use the quadratic equation to solve for a, b and c.

Answer 5

You have a+b=6, 3ab=c
3ab=c => ab = c/3

Also, (a+b) = 6 => (a+b)^2 = 36 => a^2 + b^2 +2ab = 36
Substituting ab in this we get
a^2 + b^2 +(2/3)c = 36
=>(2/3)c = 36 - a^2 - b^2
=>c = (3/2)*(36 - a^2 - b^2)

Answer 6

majdi b is right, that there is not enough information to find C.

Example:
If all we know is a + b = 6 and 3ab = C, then we could have a=5, b=1, c=15

OR we could have a=2, b=4, c = 24
OR lots of other solutions.

Answer 7

The quadratic formula is for a second degree equation in one unknown. You can not use it in this situation. With three unknowns, you will need to have three equations to solve the system. You don't have enough information.

Answer 8

I examine the formula as c=(a million/5)(q)(i) that's easy, you may isolate q You do this by "shifting" (a million/5)(i) to the different area of the equation. You do this by multiplying the two facets by its inverse or 5/i subsequently 5(c)/(i)=q or q=5c/i

Answer 9

u can not resolve it: u have 3 unknown and only 2 equations, u need a third equation to find a solution

Answer 10

3ab=15
3a(6-a)=15
18a-3a^2=15
6a-a^2=5
only 1 option .. a must be 1
so,
6-1=5
and thus b=5