Solving a Quadratic with the Quadratic Formula

Screens are sold by one measurement, and it is the one nobody measures at home: corner to corner. Nobody quotes you the width. So to know whether a screen thirteen centimetres across the diagonal fits the slot in your bag, you have to work the width out from the diagonal and the shape. Do that, and the width comes out at seven point six centimetres. That is not a number any pair of brackets will ever hand you. Real measurements come out untidy, and untidy is where factorising runs out of road. One method solves every quadratic equation, tidy numbers or not. The whole difficulty in using it is the minus signs, and that is what this video is about.

Here is the tool itself, and then where its three numbers come from.

One note first. The quadratic formula is Higher tier only. If you are sitting Foundation it is not on your paper, and the video called Solving x squared plus b x plus c equals zero by Factorising is the one you want.

The formula says this. x equals minus b, plus or minus the square root of b squared minus four a c, all over two a. It was not invented out of nowhere. It is what completing the square on a x squared plus b x plus c gives you in general, done once so that nobody has to do it again. It only works on an equation in one exact shape: a x squared, plus b x, plus c, equals zero. Getting an equation into that shape is the job of the video called Rearranging and Factorising a General Quadratic Equation. Everything here arrives already in shape. Look hard at that shape, because the trap hides inside it. The template is built out of plus signs: plus b x, plus c. So when your equation says minus five x, that minus has nowhere to go except inside the letter. b is not five. b is minus five. The sign belongs to the number, not to the gap in front of it.

So here is your handle for this video, and say it out loud. Sign with the number, number in brackets. Read a, b and c off the equation with their signs attached, then write each one in wearing its own brackets.

Quick check, and it is the whole skill in one step. In two x squared minus five x minus three equals zero, what is b? Five, minus five, or minus five x?

Take your pick. I'll wait. The answer is minus five. Not five, because the sign travels with the number it belongs to. And not minus five x, because b is only the coefficient, the number multiplying the x. Run that rule across the whole equation and a is two, b is minus five, and c is minus three.

Now the substitution, one piece at a time, because this is the step that quietly goes wrong. All three numbers go into the formula inside brackets. So minus b becomes minus, open bracket, minus five, close bracket. Minus a negative is a positive, so the top starts at plus five. Skip the brackets and you end up writing minus five where plus five belongs, and every single number after that point comes out wrong. That one small pair of brackets is doing all of the work here. Next, b squared. b is minus five, so that is bracket minus five bracket, all squared, which is a positive twenty five. The minus sits inside the bracket, so it gets squared along with the five. Written without the bracket, minus five squared means square the five first and then apply the minus, and you land on minus twenty five. Squaring happens before the minus sign unless a bracket says otherwise. Then minus four a c. a is two and c is minus three, so that is minus four, times two, times minus three, which comes to plus twenty four. Under the root, twenty five plus twenty four is forty nine, and its square root is exactly seven. The bottom is two a, which is four. That leaves x equals five, plus or minus seven, all over four. Plus or minus means two answers, so split the line. Five plus seven is twelve, over four, which is three. Five minus seven is minus two, over four, which is minus nought point five. So that equation has two solutions, x equals three and x equals minus nought point five, and both came out of one careful substitution.

Your turn now. Same shape of equation, different numbers, nothing new to learn first. Solve three x squared minus two x minus five equals zero with the formula. Signs attached, everything in brackets, and give me both answers. Pause it there and work it out. I'll wait. Here we go. a is three, b is minus two, and c is minus five. Minus b is plus two. b squared is bracket minus two bracket squared, which is four. And minus four a c is minus four, times three, times minus five, which is plus sixty. Four plus sixty is sixty four, and its square root is eight. The bottom is two times three, which is six. So x equals two, plus or minus eight, all over six. Two plus eight is ten over six, which is five thirds. Two minus eight is minus six over six, which is minus one. Five thirds is a fine final answer, or one point six seven to two decimal places.

Next, the version a real exam question asks, where the equation has to be built before it can be solved. A rectangle has a length of x plus three centimetres and a width of x centimetres. Its diagonal is thirteen centimetres. Find x, to one decimal place. The diagonal cuts the rectangle into two right angled triangles, so Pythagoras applies. x squared, plus bracket x plus three bracket squared, equals thirteen squared, which is one hundred and sixty nine. Expanding gives two x squared plus six x minus one hundred and sixty equals zero, and halving every term leaves x squared plus three x minus eighty equals zero. That tidying up is the rearranging skill taught in the video called Rearranging and Factorising a General Quadratic Equation. Now a is one, b is three, and c is minus eighty. Minus b is minus three, b squared is nine, and minus four a c is plus three hundred and twenty. Nine plus three hundred and twenty is three hundred and twenty nine, which is not a perfect square. Exactly, then, x equals minus three, plus or minus the square root of three hundred and twenty nine, all over two, and written like that it is already complete and correct. Exact or decimal is not your decision. The question decides, and this one asked for one decimal place, so out comes the calculator, and the rounding happens once, at the very end. The square root of three hundred and twenty nine is about eighteen point one four. Minus three plus that, over two, gives seven point six. Minus three minus that, over two, gives minus ten point six. And now a step that gets its own name here. The sense check. x was a width in centimetres, and a rectangle cannot be minus ten point six centimetres wide. That root is not bad arithmetic. It is a real solution to the equation that the shape cannot use, so you reject it and you say so on the page. The answer is x equals seven point six centimetres. The formula answers the equation. Only you can answer the question, and those two are not always the same thing.

There are two places this comes apart under exam conditions, and both of them are on record. Here is one line from an examiner report, about a question where the quadratic formula was the method needed. Many students did not know that the solutions needed the application of the quadratic formula. Those who did, often could not remember the correct formula. Sign errors were made when rearranging the equation and there were substitution errors when the correct formula was used. Three failures in one report, and only the first is about not knowing the formula. The other two are sign errors and substitution errors, which is precisely the step we have been slowing down. The brackets are not fussiness. They are the whole defence against what that report describes. The second place it comes apart is the last line of your answer. This next report describes a question with a geometrical context, where one of the two solutions had to be discarded. Part (b) was misinterpreted in many responses which failed to link the final answer to the original question, thereby writing nine and minus ten on the answer line, instead of nine only and discarding the negative solution because of the geometrical context of the question. Failed to link the final answer to the original question. Both roots were correct, and the answer line was still wrong, because nobody read it back against the shape being described. So write both roots down, then read the question one more time, and cross out the one the situation cannot hold.

Say that rule with me one more time, and then we will run the whole method back from the top. Sign with the number, number in brackets. Read a, b and c off the equation with their signs, then write all three in inside brackets, so minus b turns into plus five when b is minus five, and b squared stays positive. Work out the part under the root first, take the square root, then split the plus or minus into two separate answers. Leave the answer exact or round it, whichever the question asked for, and round once at the very end. Then run the sense check on what you have found. If the x you solved for stands for a length, a price, or a number of people, throw away the root that could not possibly be one of those things.

Next in the chapter: Solving Simultaneous Linear Equations by Elimination, where two equations get solved together.

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