math help can ya?

Question

A student says that the equation (a-9)(a-8)= 20 can be solved as follows : a-9=4 or a-8=5. If this is correct then explain why! I not explain why.How would u advise this student?

Answer 1

[02]
4*5=20,there is no doubt about it.But so is 1*20,2*10,5*4 etc etc.And there would be infinite no of factors if decimal or fractions are taken into account
Hence it is not possible to say with certaininty that a-9=4 and a-8=5.Therefore.the solution in that way is invalid

Answer 2

The way to solve the equation is as follows:
(a-9)(a-8) = 20
a^2 - 17a + 72 = 20
a^2 - 17a + 52 = 0
(a-13)(a-4) = 0
a = 13 or 4

Coincidently, using the theory of the student would give you a=13, but it doesn't give you a=4 (and both are solutions to the problem). But the student could have easily said a-9 = 10 and a-8 = 2 (meaning a could be 19 or 10, neither of which is correct).

Answer 3

Hmm....technically I think what the student is saying is true, but a very odd way of solving the problem.

From the logical standpoint
a-9=4 AND a-8=5 must be true for it to be 20

but
a-9=4 if a-8=5 and vice versa, and so if one of them is true, the other is automatically true. I don't know how I would advise this poor student.

Edit: right, negative numbers were not considered. (-4)(-5) would also work, so a-9 = -5 where a=4 would also work.

Answer 4

No it is not correct. You cannot assume a-9 = 4, it could equal 5 or 2 or 10 because they are all factors of 20.

The only reason when we say (x-a) = 0 and (x-b)=0
when we have (x-a)(x-b)=0 is because anything X by 0 is 0

to solve the eq, you have to expand and factorize

a^2 - 17a +72 = 20
a^2 - 17a + 52 = 0

then factorize it out or use the quad formula

Answer 5

The way to solve the problem is to expand (a-9)(a-8) -> a^2-17a+72=20, then solve for a: a^2-17a+52=0, then use the quadratic eqation to find a=4 and a=13.
Or simply factor a^2-17a+52 into (a-4)(a-13).

Answer 6

(a-9)(a-8) = 20
a² - 17a + 72 = 20
a² - 17a + 52 = 0
(a - 13)(a - 4) = 0
a = 13 or a = 4

if a = 13, then a-9 = 4 and a-8 = 5, so the student is correct.

if a = 4, then a-9 = -5 and a-8 = -4, and that's another solution.

Answer 7

You might get the right answer, but this is not the correct way to do the problem. Here is how:
(a-9)(a-8) = a^2 -17a + 72 = 20
a^2 - 17a + 72-20 = 0
a^2 - 17a + 52 = 0
(a-13)(a-4) = 0
a = 13, 4
So I guess that you do not get the correct answer that way after all.

Answer 8

tats not rite...its right only wen the right hand side is equal to 0 like this:(a-9)(a-8)=0.Then u can write as a-9=0 and a-8=0.Rite side u hav a constant....u cant equate simply by factorising.Then y do u split it as 5 &4.There are chances like 20,1;10,2....so its wrong