Quadratic functions

In this video, we’re going to look at quadratic functions. So we’re going to factorise and solve a quadratic equation. So we want to factorise x squared plus 4x subtract 12, first of all. Well, to factorise it, it would be in the form x plus a multiplied by x plus b. And if we multiply out this bracket, we would get x squared plus ax plus bx plus ab. Which can be rewritten as the following. So we can see that we need ab to be equal to negative 12 and a plus b to equal to 4. So if we write down the factors of negative 12, we would get the following factor pairs.

And we need the factor pair that adds up to make 4, which is negative 2 plus 6. And so we can write x squared plus 4x subtract 12 as x subtract 2 multiplied by x plus 6. So we now want to solve the equation x squared plus 4x subtract 12 is equal to 0. So using our answer from part A, we can write x subtract 2 multiplied by x plus 6 is equal to 0. This means either x subtract 2 is equal to 0, which is when x is equal to 2, or when x plus 6 is equal to 0, which is when x is equal to negative 6.

So instead, you can use the quadratic equation formula, which is when we have an equation of the form ax squared plus bx plus c equals 0. We can solve it using x is equal to negative b, add or minus the square root of b squared, subtract 4 times a times c, all over 2a. So if we want to solve 5x squared subtract 3x equals 2, we need this right hand side to be equal to 0. So we subtract 2 from both sides. And then, we can see a is equal to 5, b is equal to negative 3, and c is equal to negative 2. So we put these into the quadratic formula.

And we have to, of course, be careful with negative numbers here, when doing the calculations. What we will get, that this will become in the following form, and we know the square root of 49 is 7. So we’ve got two solutions. When we do the plus, this is when x is 3 plus 7 divided by 10, which is 1. Or the other solution is when 3 subtract 7 over 10, which is negative 0.4. So we have either x is 1 or x is negative 0.4. We can also use the quadratic equation formula to solve other equations.

So we might have an equation of the form a multiplied by some function of x squared, plus b times the same function of x, plus c, equals 0. And we can use the quadratic equation to solve what f of x is. So we want to solve the equation x to the power of 6 subtract x cubed subtract 100 equals 0. So we can use the fact that x to the power of a all to the power of b is equal to x to the power of a multiplied by b to put this in the form x cubed squared, subtract x cubed, subtract the 100 equals 0. Such that in our equation, f of x would be x cubed.

So we can solve what x-cubed is equal to using the quadratic equation, with a is 1 b is negative 1, and c is negative 100. So if we put this into this formula, we have the x cubed is negative b add or minus b squared, subtract 4, multiplied by a, multiplied by C, all divided by 2 times a. And we keep simplifying this until we get that x cubed is of the following form. So if we take first the plus solution, that means x cubed is 1 plus the square root of 401 over two, which we can take the cubed root to get the x to decimal places will be 2.19.

And similarly, for the minus solution, we have 1 subtract the square root of 401 divided by 2. This is x cubed. So we take the cubed root of this to get that the solution of x to two decimal places is negative 2.12. So we found our two solutions to that equation. Finally, we’re going to look at the discriminant. So the discriminant is the expression that b squared subtract 4 times a times c that appears in the quadratic equation formula. So it’s the square root part of the quadratic equation. And if the discriminant is greater than 0, then the equation has two real solutions. If the discriminant is less than zero, the equation has no real solutions.

And if the discriminant is equal to 0, then the equation has just one real solution. So for each of the 3 following equations, we’re going to find how many solutions there are. So for a, we have 4x squared subtract 12x plus 9. So a is 4, b is negative 12, c is 9. And when we put these numbers into the discriminant, we get that discriminant is equal to 0. Which tells us, according to the pink box, if the discriminant equal to zero, the equation has one solution. For b, it’s not quite in the format we need. We need the right-hand side to be equal to 0. So we’re going to add 2x to both sides.

Then, we have a is 1, B is 2, and C is 5. And putting this into the discriminant, we get that the discriminant is negative 16, which is less than 0. So the equation has no real solutions. Finally, for C, we have a is 3, b is negative 3, and c is negative 17. So the discriminant, when we put them numbers in, is equal to 213, which is greater than 0. So the equation has two real solutions. So in the comments below this video discuss the following. We want to find the values of k for which the quadratic equation below has real solutions.

So either discuss one step of the solution or a common mistake that you think might occur when solving this question. Thank you for listening.