Nine Times Ten Is Ninety

Algebra Level 2

$\large n(n+1) = 9 \times 10$

Is it true that the only solution is $n=9$ ?

Yes, it is true No, it is not true

3 solutions

Viki Zeta
Sep 6, 2016

$n(n+1) = 9 \times 10 \\ n^2 + n = 90 \\ n^2 + n - 90 = 0 \\ n^2 + 10n - 9n - 90 = 0\\ n(n+10)-9(n+10) = 0 \\ (n-9)(n+10) = 0 \\ n = 9, n = -10$

Therefore, it has 2 real solutions $9, -10$

Typo: $10$ is not a solution, but $-10$ is.

Jesse Nieminen - 4 years, 9 months ago

Yeah thanks for informing.

Viki Zeta - 4 years, 9 months ago

Also, you should use word "has" instead of "have" since "has" is for singular 3rd person. :P

Jesse Nieminen - 4 years, 9 months ago

-10 is too negligible to positive it can be neglect so answer could be 9 which is true.

A Former Brilliant Member - 4 years, 9 months ago

I don't quite understand your comment. Of course $-10$ is not negligible because it is a solution.

Jesse Nieminen - 4 years, 9 months ago

Oh,thats what i was thinking.

A Former Brilliant Member - 4 years, 9 months ago

-10 is never negligible. It is alast a real number. Which contradicts the statement in problem

Viki Zeta - 4 years, 9 months ago

How do you delete the n and the 9 in the fifth line?

Francisco Ramirez - 4 years, 9 months ago

no i dont get you?

Viki Zeta - 4 years, 9 months ago

Munem Shahriar
Oct 5, 2017

$n = -10$ is a solution too.

$(-10) (-10 + 1)$

$= (-10)(-9)$

$= (10)(9)$

$= 90$

Hence it is $\boxed{ \color{#20A900}\text{No, it is not true}}$

Elysa Picorelli
Sep 13, 2016

You could also assume 3^2 = n, which is technically 9, but not literally.