# Math Help - Quadratics - sequences and squares

1. ## 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

2. Hello, Soriku Strife!

Write a formula for the $n^{th}$ 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: . $f(n) \;=\;an^2 + bn + c$
. . and we must determine $a,\,b,\,c.$

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

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

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

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

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

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

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

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

3. Thanks, made more sense than the way I did it

4. Hello again, Soriku Strife!

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

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

To complete the square: take one-half of the x-coefficient and square it.

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

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

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

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

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

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