# Finding a question related to 'finding the square' difficult

• Aug 3rd 2012, 09:47 PM
DonGorgon
Finding a question related to 'finding the square' difficult
I'm currently partaking a distance learning course in A-level mathematics, and have reached chapter 4 in Pure Core Maths 1&2. Found everything pretty straight-forward so far, yet this question I would greatly appreciate help with.

Express 3x² - 12x + 5 in the form A(x - B)² - C :

From the first part of the question I have : 3(x - 2)² -7

I'm then asked to find the minimum value of each of the following expressions:

1. 3x² - 12x + 5 = -7 (The value of x obviously being 2).

And 2. (3x² - 12x +5)²

It's part 2 that's confusing me? Really unsure of where to start.
• Aug 3rd 2012, 10:16 PM
Prove It
Re: Finding a question related to 'finding the square' difficult
Quote:

Originally Posted by DonGorgon
I'm currently partaking a distance learning course in A-level mathematics, and have reached chapter 4 in Pure Core Maths 1&2. Found everything pretty straight-forward so far, yet this question I would greatly appreciate help with.

Express 3x² - 12x + 5 in the form A(x - B)² - C :

From the first part of the question I have : 3(x - 2)² -7

I'm then asked to find the minimum value of each of the following expressions:

1. 3x² - 12x + 5 = -7 (The value of x obviously being 2).

And 2. (3x² - 12x +5)²

It's part 2 that's confusing me? Really unsure of where to start.

Are you allowed to use Calculus?
• Aug 4th 2012, 03:06 AM
DonGorgon
Re: Finding a question related to 'finding the square' difficult
I haven't covered any calculus yet nope.
• Aug 4th 2012, 03:17 AM
Prove It
Re: Finding a question related to 'finding the square' difficult
Quote:

Originally Posted by DonGorgon
I haven't covered any calculus yet nope.

Hm, well first notice that \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 \end{align*}, being a square, can never be negative. So it makes sense that the minimum value would be 0. Let's try to evaluate where the function will be 0...

\displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 &= 0 \\ 3x^2 - 12x + 5 &= 0 \\ x &= \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)} \\ x &= \frac{12 \pm \sqrt{144 - 60}}{6} \\ x &= \frac{12 \pm \sqrt{84}}{6} \\ x &= \frac{12 \pm 2\sqrt{21}}{6} \\ x &= \frac{6 \pm \sqrt{21}}{3}\end{align*}

So the minimum value is 0, which occurs when \displaystyle \begin{align*} x = \frac{6 - \sqrt{21}}{3} \end{align*} or \displaystyle \begin{align*} \frac{6 + \sqrt{21}}{3} \end{align*}
• Aug 4th 2012, 03:47 AM
DonGorgon
Re: Finding a question related to 'finding the square' difficult
Quote:

Originally Posted by Prove It
Hm, well first notice that \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 \end{align*}, being a square, can never be negative. So it makes sense that the minimum value would be 0. Let's try to evaluate where the function will be 0...

\displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 &= 0 \\ 3x^2 - 12x + 5 &= 0 \\ x &= \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)} \\ x &= \frac{12 \pm \sqrt{144 - 60}}{6} \\ x &= \frac{12 \pm \sqrt{84}}{6} \\ x &= \frac{12 \pm 2\sqrt{21}}{6} \\ x &= \frac{6 \pm \sqrt{21}}{3}\end{align*}

So the minimum value is 0, which occurs when \displaystyle \begin{align*} x = \frac{6 - \sqrt{21}}{3} \end{align*} or \displaystyle \begin{align*} \frac{6 + \sqrt{21}}{3} \end{align*}

Thanks for getting back to me. Ah ok, from my attempt I used the expression in the form A(x-B)² - C :

3(x - 2)² - 7 = 0

3(x - 2)² = 7

3(x - 2) = ± √7

(x - 2) = ± √7 / 3

x = 2 ± √7 / 3
• Aug 4th 2012, 04:10 AM
Prove It
Re: Finding a question related to 'finding the square' difficult
Quote:

Originally Posted by DonGorgon
Thanks for getting back to me. Ah ok, from my attempt I used the expression in the form A(x-B)² - C :

3(x - 2)² - 7 = 0

3(x - 2)² = 7

3(x - 2) = ± √7

(x - 2) = ± √7 / 3

x = 2 ± √7 / 3

By the order of operations, exponentiation is done before multiplication. So when you go in reverse and solve for x, you need to undo the multiplication first.