Thread: 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.

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?

I haven't covered any calculus yet nope.

Originally Posted by DonGorgon

I haven't covered any calculus yet nope.

Hm, well first notice that , 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...



So the minimum value is 0, which occurs when or

Originally Posted by Prove It

Hm, well first notice that , 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...



So the minimum value is 0, which occurs when or

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

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.