Can you explain the solution to this Completing The Square related question please?

I have the solution (at bottom of this message).

I know how to solve the equation (part 2 of the question), but I don't understand how you would go abnout solviong it and looking art the solution, I don't get why you would do each section and what each section is doing. I would never be able to do it in an exam as I don't understand the method.

Please can you explain it to me? (I know what is meant by p, q, and r when completing the square).


Thanks, appreciate it!

I have the solution (at bottom of this message).

I know how to solve the equation (part 2 of the question), but I don't understand how you would go abnout solviong it and looking art the solution, I don't get why you would do each section and what each section is doing. I would never be able to do it in an exam as I don't understand the method.

Please can you explain it to me? (I know what is meant by p, q, and r when completing the square).

The wanted form has been expanded fully in terms of p, q and r. Then we know the coefficients of $x^2, x$ and the constant must be the same in order for the equation to hold. So you make up some equations by considering each coefficient individually and solve for them.

For the expressions to be equal for all values of x, they have to be the same expression. You therefore expand the RHS to get a quadratic in x and set each of the coefficients equal to the same one on the LHS.

How do we know the coefficients of x^2 and x?

And what do you mean by you make up some equations by considering each coefficient individually?

I get that we know p=3, (because 3x^2), and you expand the RHS to get a quadratic. But then why do you need to get that quadratic? What does it mean by stuff like "Comparing x: 2pq = 12"?

I don't have a clue really what each step of the solution means and why you do each step.

Because on the LHS they are 3, 12 and 5 respectively therefore they must be 3, 12 and 5 on the RHS too.



I mean that once you have the RHS in the form of $Ax^2+Bx+C$ (which is the same form as the LHS) you can compare the coefficients between LHS and RHS

So $3=A$. $12=B$ and $5=C$ for whatever your A, B and C are in terms of p, q and r. So then you can solve for p, q and r.


Again, coefficients must be the same so the coefficient of x must be the same on both sides hence 2pq=12.



Time to learn something new then.

I'm guessing that we know pq^2 + r is the constant because there is no x in there? And the form of a quadratic is ax^2 + bx + c, so it is the constant, c, because there is no x in that part of the equation? So basically wherever there isn't an x, that is the constant?

Also, when you say the "coefficients must be the same so the coefficient of x must be the same on both sides, hence 2pq=12", what do you mean when you say the coefficient of x must be the same? Do you mean the bit before x, so 2pq(x) and 12(x)?