Find values for *a*, *b* and *c*, so that *x* = 3:

Later, a second formula will be provided if you haven’t already got it.

Hint: none of the four digits are outside the range minus 9 to plus 9.

Here’s that formula:

Find values for *a*, *b* and *c*, so that *x* = 3:

Later, a second formula will be provided if you haven’t already got it.

Hint: none of the four digits are outside the range minus 9 to plus 9.

Here’s that formula:

Looks like a quadratic. So (x-3) squared. a=1, b=6, c=9

Sorry b= -6

Not thinking straight

That quadratic formula instantly rang a bell, a bell which had previously been un-rung for more than 50 years – I couldn’t be arsed then and can’t be arsed now.

As it didn’t seem to offer a certain channel to a Littlewoods jackpot or the chance of a first leg-over from the neighbouring girls’ school, it never seemed an important enough use of my precious teenage time.

Me brain ‘urts.

x= +3a= +2b= -4c= -6I agree with Mark in Mayenne. (x-3)*(x-3) = 0. This has the single solution x = +3.

James’s solution factorises as (x-3)*(2x+2) = 0. This has two different solutions for x; x = +3 and x = -1.

If one allows solutions with a second different value for x than +3, there are an infinite number of solutions.

Best regards

Last para – but one doesn’t, as plus 3 was the given the question was based upon.

In addition to my earlier point (which James has clearly not understood), James’s solution ([a,b,c]=[+2,-4,-6]) has an arbitrary factor of 2 in it. Normal mathematical processing is to cancel out the common constant (2 in this case) giving [a,b,c]=[+1,-2,-3] as a solution.

Just to repeat, the correct solution for x = +3 is [a,b,c]=[+1,-6,+9]. Otherwise there are an infinite number of solutions: (x-3)(x-n)=0 where n can take any value.

Best regards

He actually did understand, hence the wording:

Find values for… rather than the

onecorrect answer.However, there aren’t all that many when they must be whole numbers between minus 9 and plus 9.