How do you do this question from a maths paper?

Question

For all values of x and m, x^2 – 2mx = (x – m)^2 – k

(a) Express k in terms of m

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(b) The expression x^2 – 2mx has a minimum value as x varies.

(i) Find the minimum value of x^2 – 2mx

Give your answer in terms of m.

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(ii) State the value of x for which this minimum value occurs.

Give your answer in terms of m.

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Thanks in advance :)

Answer 1

(a)  Solve for k:

x^2 – 2mx = (x – m)^2 – k.  Get k by itself on one side of the equal sign by adding k to both sides, and then by subtracting (x^2 - 2mx):

x^2 - 2mx + k = (x-m)^2 - k + k

x^2 - 2mx + k = (x-m)^2

-x^2 + 2mx + x^2 - 2mx + k = (x-m)^2 - x^2 + 2mx

k = (x-m)^2 - x^2 + 2mx.  (x-m)^2 = (x-m)(x-m)

k = (x-m)(x-m) - x^2 + 2mx. FOIL:

k = x^2 - mx - mx + m^2 - x^2 + 2mx.  

k = x^2 - 2mx + m^2 - x^2 + 2mx.  You can see that all of the terms with x cancel each other out.  You're left with:

k = m^2

(b)

(i)  Okay, so you have the equation x^2 – 2mx = (x – m)^2 – k, and you now know what k equals.  Plug the value of k into the equation:

x^2 – 2mx = (x – m)^2 – k

x^2 – 2mx = (x – m)^2 – m^2

The expression (x-m)^2 will ALWAYS be greater than or equal to zero, right? It cannot be negative because (x-m) is squared.

So if (x-m)^2 were zero, what would (x^2 - 2mx) be equal to?

x^2 – 2mx = (x – m)^2 – m^2

x^2 – 2mx = 0 – m^2

x^2 – 2mx = -m^2

So the minimum value is -m^2.

(ii)  Solve for x from the equation above, x^2 – 2mx = -m^2:

x^2 – 2mx = -m^2.  Add m^2 to both sides to get...

x^2 - 2mx + m^2 = 0.  Factor by grouping:

(x - m)(x - m) = 0.

(x - m)^2 = 0.  Take the square root of both sides to get...

x - m = 0.  Add m to both sides to get...

x = m

The minimum value occurs when x = m.

I hope that helped!