# Quadratics - sequences and squares

• Oct 3rd 2007, 11:13 AM
Soriku Strife
Quadratics - sequences and squares
Okay, I've got this question and I've spent about an hour trying to figure it out and it's doing my head in:

Write a formula for the nth term of this quadratic sequence:
7, 15, 27, 43, 63
nth term = ??

Write this expression in completed square form:
x² - 10x +2

Hence solve
x² - 10x +2 = -5

Ok, any help would be appreciated as I've tried many ways of doing it....
Thanks =D
• Oct 3rd 2007, 12:28 PM
Soroban
Hello, Soriku Strife!

Quote:

Write a formula for the $\displaystyle n^{th}$ term of this quadratic sequence:
. . 7, 15, 27, 43, 63, . . .

Here's one approach to this problem (bear with me, it's quite long).

We write a general quadratic function: .$\displaystyle f(n) \;=\;an^2 + bn + c$
. . and we must determine $\displaystyle a,\,b,\,c.$

We have three "unknowns", so we need three equations.
We'll use the first three terms of the sequence.

The first term is 7. .That is: .$\displaystyle f(1) = 7$
. . $\displaystyle a\cdot1^2 + b\cdot1 + c \:=\:7\quad\Rightarrow\quad a + b + c \:=\:7\;\;{\color{blue}[1]}$

The second term is 15. .That is: .$\displaystyle f(2) = 15$
. . $\displaystyle a\cdot2^2 + b\cdot2 + c\:=\:15\quad\Rightarrow\quad 4a + 2b + c \:=\:15\;\;{\color{blue}[2]}$

The third term is 27. .That is: .$\displaystyle f(3) = 27$
. . $\displaystyle a\cdot3^2 + b\cdot3 + c \:=\:27\quad\Rightarrow\quad 9a + 3b + c \:=\:27\;\;{\color{blue}[3]}$

$\displaystyle \begin{array}{ccccc}\text{Subtract {\color{blue}[1]} from {\color{blue}[2]}:} & 3a + b & = & 8 & {\color{blue}[4]} \\ \text{Subtract {\color{blue}[2]} from {\color{blue}[3]}:} & 5a + b & = & 12 & {\color{blue}[5]}\end{array}$

Subtract [4] from [5]: .$\displaystyle 2a \:=\:4\quad\Rightarrow\quad\boxed{ a \:=\:2}$

Substitute into [4]: .$\displaystyle 3(2) + b \:=\:8\quad\Rightarrow\quad\boxed{ b \:=\:2}$

Substitute into [1]: .$\displaystyle 2 + 2 + c \:=\:7\quad\Rightarrow\quad\boxed{ c \:=\:3}$

Therefore, the function is: .$\displaystyle f(n) \;=\;2n^2 + 2n + 3$

• Oct 3rd 2007, 12:32 PM
Soriku Strife
Thanks, made more sense than the way I did it :)
• Oct 3rd 2007, 01:43 PM
Soroban
Hello again, Soriku Strife!

Quote:

Write this expression in completed square form: .$\displaystyle x^2 - 10x + 2$

We have: .$\displaystyle x^2 - 10x + 2$

To complete the square: take one-half of the x-coefficient and square it.
Add and subtract this quantity.

. . $\displaystyle x^2 - 10x \,{\color{blue}+\,25} + 2 \,{\color{blue}-\,25}\;=\;(x^2-10x + 25) - 23 \;=\;(x-5)^2-23$

Quote:

Hence, solve: .$\displaystyle x^2 - 10x +2 \:= \:-5$

We have the equation: .$\displaystyle (x-5)^2 - 23 \:=\:-5$

Add 23 to both sides: .$\displaystyle (x - 5)^2 \:=\:18$

Take square roots: .$\displaystyle x - 5 \;=\;\pm\sqrt{18}\quad\Rightarrow\quad x - 5\;=\;\pm3\sqrt{2}$

Add 5 to both sides: .$\displaystyle \boxed{x \;=\;5 \pm3\sqrt{2}}$