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Video: Finding Roots of Quadratics by Factoring

Tim Burnham

Catch on how to find the roots of a quadratic by factoring a quadratic expression. Examples include x^2 + 10x = 0, (x + 7)^2, x^2 + 2x - 35 = 0, 4x^2 - 25 = 0, and 6x^2 + 11x - 10 = 0. Also, it covers different ways of presenting answers, such as solution sets.

17:25

Video Transcript

In this video, we’re gonna look at some examples of quadratic equations and use factoring, or factorizing, to find their roots. We’ll also talk about some of the different notations that you can use to present your answers. We’ve covered factoring different types of quadratics in more detail in other videos, so we’ll just quickly recap the process here.

But first, remember that a quadratic expression has a squared term, a linear term, and a constant term. So that’s a positive or negative number times 𝑥 squared, a positive or negative number times 𝑥, and a pogi- positive or negative constant on the end. Also remember that the coefficient of 𝑥 or the constant term on the end, so the 𝑏 or the 𝑐 value here, could either be zero or they could both be zero for that matter. But the 𝑎-value here can’t be zero, or otherwise it wouldn’t be a quadratic.

So for example if 𝑎 was one, 𝑏 was negative five, and 𝑐 was six, we’d have a quadratic expression of 𝑥 squared minus five 𝑥 plus six. And also remember it doesn’t have to be 𝑥; it could be any letter. The roots of a quadratic are the 𝑥-values which generate your result of zero, when you substitute them into the expression. So for example a question like: Find the roots of the quadratic equation 𝑦 equals 𝑥 squared minus five 𝑥 plus six means solve 𝑥 squared minus five 𝑥 plus six equals zero. In other words, find the values of 𝑥 which match that equation.

Now in this case, let’s not worry for now how we found them but there are two possible answers. When 𝑥 is equal to two, we put that number in and we get an answer of zero, so that works. And when 𝑥 is equal to three, we put that number in for 𝑥 and we get an answer of zero, so that works as well.

Now we can present our answer in two ways. We can either present it as a list, so 𝑥 is equal to two or 𝑥 is equal to three; or we can present it as a solution set, so two and three in set notation. And with quadratics you are likely to come across this idea of real numbers, so our solution set is from the set of real numbers. So thinking back to what our quadratic graphs look like, if the graph cuts the 𝑥-axis in two places, then we’ve got two solutions. If it touches the 𝑥-axis in one place, then we’ve got one solution. And if it doesn’t cut through the 𝑥-axis at all, then we’ve got no real solutions. If we use imaginary or complex numbers then we can come up with some solutions, but we’re not gonna delve into that right now.

Okay. Let’s look at some examples. So number one: Find the solution set of 𝑥 squared plus ten 𝑥 equals zero. This might also have said: Find the roots of 𝑦 equals 𝑥 squared plus ten 𝑥.

Now this is an equation which easily factors or factorizes. We’ve got 𝑥 and 𝑥 in here, so 𝑥 will be our common term that we can take out. So we can express this as 𝑥 lot- times 𝑥 plus ten is equal to zero. Now that 𝑥 right up against the bracket there means it’s 𝑥 times 𝑥 plus ten, so we’ve got something times something is equal to zero. Now the only way you can get an answer of zero when you multiply two things together, is if one of those is zero. So either 𝑥 is equal to zero, or 𝑥 plus ten is equal to zero. And if I subtract ten from each side of that equation, I’m left with 𝑥 is equal to negative ten. So if I was just writing those as a list, I would say my answer is 𝑥 equals zero or 𝑥 equals negative ten. But the question has asked for a solution set. So using set notation, the two values that 𝑥 can take are zero and negative ten; so this would be my answer.

Number two: Find the solution set of 𝑥 plus seven all squared is equal to zero.

Now 𝑥 plus seven all squared means 𝑥 plus seven times 𝑥 plus seven. So we’ve got something times something is equal to zero. Again something times something equals zero, the only way we can get that is if one of those things is equal to zero itself. So either 𝑥 plus seven is equal to zero, which would make 𝑥 equal to negative seven, or the same thing. So we’ve got repeated roots. This is one of those situations where if we drew the graph of this function 𝑦 equals 𝑥 plus seven all squared, it would look roughly like that. It would touch the 𝑥-axis in one place at negative seven, and in fact it would cut the 𝑦-axis at forty-nine. So our solution set has only got one item in it, negative seven.

Moving on to number three then. Find the solution set of 𝑥 squared plus two 𝑥 minus thirty-five equals zero in ℝ, the set of real numbers.

So it’s just worth remembering that just 𝑥 squared means one 𝑥 squared. And when we’ve got one 𝑥 squared, there’s a certain set of rules that we can follow when we’re doing our factoring. So remember if we’ve got 𝑥 plus a number times 𝑥 plus another number and we multiply them out term by term, we’ve got 𝑥 times 𝑥 which gives us 𝑥 squared, and we’ve got 𝑥 times 𝑏 so that’s positive 𝑏𝑥, and 𝑎 times 𝑥 which is positive 𝑎𝑥, and 𝑎 times 𝑏 which is positive 𝑎𝑏.

Now looking at this we’ve got 𝑏𝑥 and 𝑎𝑥, so that’s factoring out the 𝑥 there; we’ve got 𝑎 plus 𝑏𝑥. So in converting between these two different forms of our expression, we’ve got, on the end we’ve got 𝑎 times 𝑏 so this number times this number. And here, as the multiple of 𝑥, we’ve got 𝑎 plus 𝑏. So I’m going to add 𝑎 to 𝑏. So that should help us to do our factoring remembering that-that format.

So I’m looking for 𝑥 plus or minus something times 𝑥 plus or minus something else. And if I multiply those two numbers together, I’m gonna get negative thirty-five and if I add them together, I’m gonna get positive two. So first of all, I’m just gonna write out all the factors of thirty-five.

So always start up with one and the number, so one times thirty-five, two isn’t a factor, three isn’t, four isn’t, five is a factor, five times seven, six isn’t a factor, and now we’re up to seven. We’ve already encountered seven in our list, so we know we’ve got all of the factors that we need; one times thirty-five or five times seven. Well they’re factors of thirty-five. Now to get negative thirty-five, one needs to be positive and the other needs to be negative. So just bare that in mind, one of these is gonna be positive, the other’s gonna be negative. Now when I add those two factors together, they’ve got to give me positive two. Now for one and thirty-five, the difference is thirty-four. So it doesn’t matter which one of those is positive and which one is negative. When I add them together, I’m never gonna get an answer of positive two. But with five and seven, their difference is two. So I’ve gotta think carefully which one has to be positive, which one has to be negative.

And when I add them together, the result is positive. So the bigger number needs to be positive and the smaller number will need to be negative. So 𝑥 minus five times 𝑥 plus seven is equal to zero. Now let’s just check that we get the same expression when we multiply those out. 𝑥 times 𝑥 is 𝑥 squared, 𝑥 times seven is seven 𝑥, minus five times 𝑥 is minus five 𝑥, and minus five times positive seven is minus thirty-five. And simplifying this down, seven 𝑥 take away five 𝑥 is two 𝑥. That leaves us with 𝑥 squared plus two 𝑥 minus thirty-five. That’s what we were looking for originally so we-we have factored it correctly.

So we find ourselves in the situation where we’ve got something times something is equal to zero. And if something times something equals zero, one of those things must be zero. So either 𝑥 minus five is zero, which would mean that 𝑥 would be five; or 𝑥 plus seven would be zero, in which case 𝑥 will be equal to negative seven. So our solution set consists of negative seven and five.

Number four: Find the solution set of four 𝑥 squared minus twenty-five equals zero in the set of real numbers.

So here’s a quadratic which has a 𝑏 value of zero. And in fact, this is a very special case because four 𝑥 squared can be written as two 𝑥 all squared, and twenty-five can be written as five squared. So we’ve got the difference of two 𝑥- of two squareds; two 𝑥 all squared minus five squared. Now let’s just take a second to remember the difference of two squareds technique. If I had two parentheses here 𝑥 minus 𝑎 times 𝑥 plus 𝑎, 𝑥 times 𝑥 is 𝑥 squared, 𝑥 times positive 𝑎 is positive 𝑎𝑥, negative 𝑎 times 𝑥 is negative 𝑎𝑥, and negative 𝑎 times positive 𝑎 is negative 𝑎 squared. So multiplying out those parentheses, I’ve got 𝑥 squared plus 𝑎𝑥 take away 𝑎𝑥. So these two terms, if I start off with 𝑎𝑥 and I take away 𝑎𝑥, that’s gonna give me zero; so that’ll cancel out. And then I’ve got minus 𝑎 squared. So I’ve got 𝑥 squared take away 𝑎 squared; the difference of two squareds. So I’m gonna use this result 𝑥 minus 𝑎 times 𝑥 plus 𝑎 gives me 𝑥 squared minus 𝑎 squared, in order to help me factor this expression here.

So something squared minus something else squared gives me the something minus the something else times the something plus the something else. So I’m gonna do the same pattern here. So this is what we get factored and remember from our original equation that that is equal to zero. So we’ve got something times something is equal to zero. So either two 𝑥 minus five is equal to zero, which when I rearrange and solve that equation I get 𝑥 is equal to five over two; or two 𝑥 plus five equals zero, which I can rearrange and solve to get 𝑥 is equal to negative five over two. So here are my two answers: 𝑥 is five over two or negative five over two. And using set notation like it asked for in the question, my solution set of real numbers is negative five over two or- and five over two.

Lastly then number five. Find the solution set of six 𝑥 squared plus eleven 𝑥 minus ten equals zero in the set of real numbers.

So here we’ve got to factorize a quadratic expression which doesn’t have one 𝑥 squared; it’s got more than one 𝑥 squared there. So we need to factor that quadratic, put it equal to zero, and see what two solutions we get. Now this is a bit trickier than those monics. So the word monic means when the coefficient of the highest power term is equal to one. So for example one 𝑥 squared plus two 𝑥 plus five. Now as we said, these monics were easier to factorize than these non monics. So let’s just go through a method then of factoring these non monic quadratics.

So first of all I’m gonna do the coefficient of 𝑥 squared, six, times the constant term negative ten. And six times negative ten is negative sixty. Then what I’m gonna do is write out all the factors of — well I’m just gonna do sixty and then we’ll deal with the minus sign later on. So one times sixty is sixty, two times thirty is sixty, three times twenty, four times fifteen, five times twelve, six times ten, seven isn’t a factor, eight isn’t a factor, neither is nine, and ten, we’ve already encountered ten; so we’ve got all of our factors of sixty.

So I’ve got six pairs of factors of sixty. But what I need to know now is which pair of those, when I add them up, do I get positive eleven; the coefficient of the 𝑥 term. Now remember because they’re gonna multiply together to make negative sixty, one of them is gotta be positive, one of them is gotta be negative. So when I add them together, I’m gonna get the difference of the two. So I’m looking for two factors that have a difference of eleven. So obviously one and sixty isn’t gonna have a difference of eleven, two and thirty isn’t gonna have a difference of eleven, nor is three and twenty, four and five [fifteen], they do have a difference of eleven. So I need to work out which one needs to be positive and which one needs to be negative. I’m trying to generate positive eleven, so the biggest one of these has to be positive and the smallest one needs to be negative. So negative four plus fifteen gives me neg- gives me positive eleven.

So I’m now gonna re-express this middle term plus eleven 𝑥 as a combination of negative four 𝑥 and positive fifteen 𝑥. Now it doesn’t matter which way around I write that, positive fifteen 𝑥 take away four 𝑥, or negative four 𝑥 add fifteen 𝑥; so I’ve chosen to do it this way. But basically, fifteen 𝑥 take away four 𝑥 gives me eleven 𝑥 like we got in the line above; so those two lines are completely equivalent. So having re-expressed that middle term, the 𝑥, I’m now gonna treat this as two separate halves, six 𝑥 squared plus fifteen 𝑥 and negative four 𝑥 take away ten 𝑥. And I’m gonna factor the first half. So six and fifteen have got a highest common factor of three, and 𝑥 squared and 𝑥 have got a highest common factor of 𝑥. So that’s gonna be three 𝑥 times two 𝑥. Cause three 𝑥 times two 𝑥 gives me six 𝑥 squared, what do I need to multiply three 𝑥 by to get fifteen 𝑥; that’s just five.

So I’ve just factored the first half of that expression up here. Now this thing in brackets, two 𝑥 plus five, I want this to be a common factor. So I’m just gonna write that bracket out again here, two 𝑥 plus five. And I need to work out what do I need to multiply that by in order to get this expression up here, negative four 𝑥 take away ten. So two 𝑥, what do I need to multiply two 𝑥 by to get negative four 𝑥. Well that would need to be negative two, negative two lots of two 𝑥 is negative four 𝑥. So let’s just check now that negative two times positive five, that gives us negative ten. Yep, that’s the other term that we’re looking for.

So these two lines are also equivalent, we’ve just rewritten it in a slightly different way. Instead of six 𝑥 squared plus fifteen 𝑥, we’ve got three 𝑥 lots of two 𝑥 plus five and instead of negative four 𝑥 take away ten 𝑥, we’ve got negative two times two 𝑥 plus five. Now I’ve got something times two 𝑥 plus five take away something times two 𝑥 plus five. That two 𝑥 plus five is a common factor to these two terms. So I’m gonna take that out as ano- as a factor. Now in the first term, I had two 𝑥 plus five times three 𝑥. And in the second term, I had two 𝑥 plus five times negative two. So I factored my quadratic expression down to two 𝑥 plus five times three 𝑥 minus two. And if we do a quick check of that two 𝑥 times three 𝑥 is six 𝑥 squared, two 𝑥 times negative two is negative four 𝑥, positive five times three 𝑥 is fifteen 𝑥, and positive five times negative two is negative ten. And as we said before negative four 𝑥 plus fifteen 𝑥 is plus eleven 𝑥. So yep, that is the expression we were looking for. So it looks like we did factorize it correctly.

So doing our factoring, we’ve got something times something is equal to zero. So one of those is gonna be zero. So either two 𝑥 plus five is zero, in other words 𝑥 is equal to negative five over two; or three 𝑥 minus two is zero, in other words 𝑥 is equal to two thirds. So our solution set is negative five over two, or two thirds. And remember if we were writing this as a list, we would’ve written 𝑥 is negative five over two, or 𝑥 is equal to two thirds.

So to summarize then, roots are 𝑥-values that substitute to give the expression a value of zero. Quadratics can have two, or one, or no real roots. Then you need to choose a method of factoring. Whether you’ve got simple factoring, difference of two squareds, a monic, or a non-monic, you can factor into two brackets. And when you put those equal to zero and you’ve got something times something equals zero, either the first thing is equal to zero, or the second thing is equal to zero. So that enables you to find your solutions. Enjoy your factoring to solve your quadratics.