I've been teaching high school math for 10 years, and every year, students in every grade are making the same mistakes. Here are the top 10 mistakes that are costing you marks and frustrating your teacher. Oh, this is an easy one. I've got negative 3 squared. I remember the trick my teacher told me.
If I've got an even exponent on a negative base, I know my answer is going to be positive. So this one... negative 3 times negative 3, that's positive 9. Perfect. The answer is actually not positive 9, you've just misinterpreted what the base of the power is. Let me make some notes over here showing you how you can correct this.
So negative 3 squared, that negative is not part of the base of the power, it's not in brackets with the 3. So negative 3 squared, you need to think of that as a negative 1 multiplied by 3 squared. which means what we have is negative 1 times 3 times 3, which is, of course, negative 9. So for that question, the base of the power is just 3. There's a negative in front of the power. What you thought it was, you thought it was this.
If the question was written like this, negative 3 in brackets squared, that would mean the base of the power is negative 3, which means we have negative 3 times negative 3, and that is positive 9. So make sure you know. This is not the same thing as this. Pay attention to what the base of the power actually is. Oh, I love it when stuff cancels out in questions. This one's easy.
I got a 2 in the numerator, a 2 in the denominator. Anything divided by itself cancels out, right? So all I'm left with is x plus 8. Easy.
No, sorry, that's not correct, but this is actually a super common mistake to actually forget to divide both of the terms by the denominator. You need to divide the 8 by 2 as well. Let me show you how to do that.
Okay, one way you could do this so that you don't make that mistake would be to separate this into two fractions being added. Since the 2x and the 8 are both being divided by 2, I could separate this into 2x over 2 plus 8 over 2, which is x plus 4. Notice the 2x and the 8 are both being divided by 2. Another way you could do this, I could common factor a 2 from the numerator, giving me 2 times x plus 4, all over 2. And now that the numerator is in factored form, you can look to cancel or reduce factors with each other. So I see a factor of 2 in the numerator, a factor of 2 in the denominator, something divided by itself is 1, so I can cancel those out, and I'm left with just x plus 4. So two different ways to think about getting that same correct answer, but x plus 8, no. Okay, I just need to simplify this. I remember distributive property.
I know you just take whatever is in front of the brackets and multiply it by everything in the brackets, and then the brackets are gone. So I've got this 9, I'll leave that there, minus, and then I have to do 5 times x, that's 5x, and 5 times negative 8, that's negative 40. And then I can collect my like terms, 9 minus 40 is negative 31, so I've got negative 5x minus 31. Alright, to start this one, I'm just going to start collecting some like terms. I've got a 9 minus 5, that's 4. It's going to be easier if I do that first. And now I'll distribute the 4 in. I've got 4x, and then minus 4 times 8, that's 32. Nice.
Okay, both of you actually just made one small mistake having to do with orders of operations. Let's correct them. Student number one over here on the left, you shouldn't have just distributed the five. You should have distributed a negative five if you were wanting to get rid of the brackets all at once. If you did it the way you did it, this product would still have to be in brackets, and you'd have to remember to subtract both of these, making this constant term wrong.
We'd be subtracting negative 40. which means plus 40, so you'd have 9 plus 40, which is 49. So you'd have negative 5x plus 49, not negative 5x minus 31. But that's not even how I would do this question. Like I said, I would distribute the negative 5, and then that would allow me to get rid of the brackets all in one step. So you'd have 9, and then negative 5 times x is negative 5x, and negative 5 times negative 8 is positive 40. So when you collect your like terms, Like I said, you'd have negative 5x plus 49. All right, what happened over here? Okay, so yeah, this student didn't follow the correct order of operations.
We can't do a subtraction first. This negative 5 is being multiplied by something. So we'd have to do the multiplication before the subtraction. So no, right away I know the answer is going to be wrong because they didn't follow the correct order of operations.
Always remember bed mass. Ooh, exponential equations. These ones are always tricky for me.
Okay, I think... Oh! Ah, this one's actually pretty easy.
I can do this one, right? 3 times 2, that's just 6. So I've got 6 to the x equals 36. And I know my powers of 6. 6 squared is 36. So my answer has got to be 2. Before I show you where the mistake in this question is, let me remind you that when solving equations, you can always check to see if your answer is right or wrong by plugging into the original equation. So if I plug 2 into the original equation, I'd have 3 times 2 squared. Does that equal 36?
Well 2 squared is 4, 3 times 4, and 3 times 4 is 12. Uh, 12 is not 36. So that shows you it's not the right answer, but what did you do wrong? You can't multiply the base of a power by a constant. That's not how it works.
So what we would have to do to solve this question is divide both sides by 3 to start off with. So we'd have 2 to the power of x equals 36 over 3, which is 12. And really what we're looking for is what exponent goes on 2 to get 12. And logarithmic functions can find us missing exponents. The exponent would equal log base 2 of 12. And you can get an approximate value for that if you wanted, but that's the exponent that would satisfy this equation.
Okay, fractions are sometimes tricky to remember all these rules, but okay, I think the first, okay, so I know bed mass, I have to do the multiplying first, so I'm going to multiply, okay, I have to do three times a fifth, okay, three times one is three, three times five is 15, okay, that looks good, plus one over two, and then adding fractions, okay, Okay, so 3 plus 1 is 4. 15 plus 2 is 17. 4 over 17. I think that's good. So you've made a couple mistakes with fraction rules here. The first of which, 3 times the 5th isn't 3 15ths.
Let me show you how you can avoid making that mistake of thinking you multiply a constant into the numerator and denominator, right? It only multiplies by the numerator. Let me show you why. So, you can think of that 3, anytime you have a whole number and you're working with fractions, I find it helps a lot if people rewrite that whole number as a fraction over 1. So that 3, think of it as a 3 over 1. So I've got 3 over 1 times 1 over 5. And when multiplying fractions, you multiply the numerators, 3 times 1 is 3, and multiply the denominators, 1 times 5 is 5. So the correct product of the first two is 3 over 5. The 3 just gets multiplied by the numerator. And when adding fractions, do not forget to get a common denominator.
One fraction has a denominator of 5, the other a denominator of 2. A common denominator between those two would be the product of them, 10. So I'll make this one be 10 by multiplying top and bottom by 2, and this one be 10 by multiplying top and bottom by 5. And that gives me 6 over 10. plus 5 over 10, and when adding fractions, you keep the common denominator, don't add the denominators like you did over here, you keep the common denominator and just add the numerator, 6 plus 5, 11. Okay, I remember my teacher saying I have to do the opposite operation to move a term between sides of the equation. Okay, on the left, I see a 3. So when I move to the right, I think I'm going to divide by 3, which means, okay, the 3 is gone from the left, which means I just have the x left. And on the right, I've got 6 over 3, which is 2. Oh, this equation over here, negative 2 times x equals 8. The opposite of multiplying by negative 2, well, the opposite of negative 2 would be positive 2, and the opposite of multiplying is dividing, so I would divide by positive 2. So I get x equals 8 over 2, x equals 4. There we go, both of those done. So you're right, you do have to do inverse operations to isolate a variable, it's just you chose the wrong operations so you got incorrect answers.
And don't forget you can check to see if your answers are right by plugging them into the original equation. If I plug 2 in for x into the first equation, 3 divided by 2 is in 6. If I plug 4 in for x into the second equation, negative 2 times 4 is not 8. Alright, let me show you how to correct the first one. To avoid making this mistake, I think you should try and approach it from more of the balanced method when solving equations.
So whatever you do to one side, do it to the other side. So when your variable's in the denominator, let's multiply both sides of the equation by that variable to get it out of the denominator. So x times 3 over x equals 6 times x. I just multiply both sides by x.
And the reason why I did that is because now we have x over x on the left, which cancel out, and we've got 3 equals 6x. Then divide both sides by 6, and I get 1 half equals x. And if we check that in the original equation, 3 divided by a half is 6. Alright, let's correct this one. I understand your thinking here, but if we divided both sides by 2, that doesn't cancel out this negative 2. Negative 2 divided by 2 is negative 1, so we'd have negative x equals 4. What we should do instead is, to the original equation, divide both sides by negative 2. The opposite of multiplying by negative 2 is dividing by negative 2. And you can see why, because these factors of negative 2 cancel out, leaving us with x equals negative 4. And if we check that in the original equation, negative 2 times negative 4 is positive 8. Oh, I love it when these easy distributive property questions show up in higher level courses.
I know how to do distributive property. I can just multiply these right into there. Sine times x plus y. Well, that's just sine times x.
plus sine times y. And same with log. Log times x plus four, that's just log times x plus log times four.
So both of these are incorrect. It's a super common mistake to think that sine and log can be treated like numbers and distribute them into their own argument. Sine and log are functions though, not numbers.
What we have with both of these aren't a product of things. It's not sine times x plus y and log times x plus four. It's sine of x plus y and log of x plus four, you can't distribute a function into its own argument.
So don't try and treat sine and log as if they were numbers and try and do distributive property sine of x plus y, it's actually fairly complicated. There's an identity for that. If you wanted to rewrite sine of the angle x plus y, it's equal to sine x cos y plus cos x sine y.
And log of x plus 4, there's nothing we could do to simplify that. You can't expand a log into its own argument. So don't try and do that.
I just have some expressions here to simplify. I remember my exponent laws. Okay, I've got things being multiplied.
I know how to multiply. Oh, I think I remember. Okay, you leave the base, yeah?
And they're being multiplied. 5 times 3, 15. This next one. Okay, this time the powers are being divided. So I keep the base. And 8 divided by 4, 2. And my last one.
Oh, and this one here. They're being multiplied. Oh, 4 times 4, that's 16. Okay, so my base is 16. And then I think I remember something about when multiplying powers, you, I think you add the exponents, 3 plus 2, 5, 16 to the 5, good.
Okay, so all three of these are incorrect. Let's start by looking at the first one. So we do have powers of x being multiplied together, and when you're multiplying powers of the same base, you keep that base of x. But you add the exponents.
5 plus 3 is 8, so we'd have x to the 8. And if you want to see why, you can take the time to write it out in factored form, right? x to the 5 means x times x times x times x times x. And we have to multiply that by x cubed, which means multiply it by 3 more factors of x.
And if you look at how many x's we have there being multiplied together, there are 8 of them. x to the 8. If we look at the second one, dividing powers with the same base, yes, you keep the same base, you did that well. But we have to subtract the exponents, not divide them.
8 minus 4, 4. And if you want to see why, write it out. So we have 8 factors of x divided by 4 factors of x. And you notice that 4 pairs will cancel out.
1 pair, 2 pairs, 3. 4. So what do we have left? In the numerator we have 4 factors of x, hence x to the 4. Okay, the last one was so close. Multiplying powers to the same base. Yes, you do add the exponents.
The exponent is 5, but you forgot what you did up here. You keep the base the same. It should be 4 to the power of 5. You don't multiply the bases together. You keep the base the same and add the exponents. Oh, solving an equation.
Easy. Inverse of squaring is square rooting. So on the other side, if I move the squared, it becomes square root.
I get the square root of nine, which is three. And let me check and see if that's right. If I plug it back in, is three squared equal to nine? Yep, nine equals nine. Good.
I know I got the right answer this time. Okay, you did get one correct answer for this question. Three is the correct answer. Three squared.
does equal 9, but it's not the only correct answer. The inverse of squaring is actually plus or minus square rooting, so you should have got plus or minus 3 as your answer, right? If you square negative 3, you also get 9, so that's also a solution to the equation.
So this is just an example how some equations can have more than one answer, so make sure you find all answers that satisfy an equation. This one. Okay, the base of my power is being squared. I can just put this exponent 2 on the x and the 3. So I've got x squared plus 3 squared. Oh, and I'll make sure I simplify that.
3 squared is 9. Perfect. Okay, so this is incorrect. You can't take the exponent and put it on both terms in the base if in the base the terms are being added or subtracted.
This is the most common mistake made in all the high school math courses. So try never to make this mistake. Look at the base of your power. If it's terms being added or subtracted, you can't just put the exponent on both.
So how can you not make this mistake? Rewrite that x plus 3 squared as x plus 3 times x plus 3. That way you can see what you need to do is you need to find the product of these two by FOILing it. I need to do the product of the x's plus the product of x times 3 plus the product of 3 times x plus the product of 3 times 3, which gives me x times x is x squared plus x times 3 is 3x plus 3 times x is another 3x plus 3 times 3 is 9. So yes, there is an x squared and a 9, but there's also a middle term. It's x squared plus 6x plus 9. I think maybe what you were thinking of when you tried to do this, like let's say it was 3 times x squared.
Yes, there is an exponent rule for that. If the base of the power is a product of two numbers, we can put the exponent on both of the factors of the base. So yes, this would equal 3 squared times x squared, which is 9x squared. So in that case, you can.
But if the base... is a sum or difference of things, you cannot do that. You have to take the time to actually expand it out. All right, now that I've gone over these common mistakes with you, I hope you never make the common mistakes that were made in this video.
Why not test yourself now? Here's a eight question true or false test. Try and figure out if each of these statements are true or false.
Post your answers down in the comments. Let's see how you do.