Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Equations reducible to quadratics. Certain equations that are not really quadratic can be thought of as quadratic like. So in number one, we know that X ^2 * X ^2 would give us X to the 4th. We also know that -1 * -9 would give us a positive 9 and add to give us -10. So X to the fourth minus ten X ^2 + 9 = 0 is really X ^2 - 1 * X ^2 - 9 = 0. Now those are both the difference of squares, so X - 1 * X + 1 and X - 3 * X + 3. If we set each of those parentheses equal to 0 and solve, we'd have X - 1 equals zero, X + 1 = 0, X -3 = 0, and X + 3 = 0. So X is 1X is -1, X is 3, and X is -3. If we look at the highest degree on this equation, it's a four. So we should expect 4 solutions, which is what we have found. We usually list them the smallest to largest, so -3 negative 1, positive one, and three. Now there is another way to do this and that's through substitution. We could say, let's let U equal X ^2. If U is X ^2 and I square each side, I'd get U ^2 equaling X to the 4th. So if U squared is X to the 4th and we substitute, we'd get U ^2 - 10, U plus 9 = 0. Now we know that that is U - 9 and U - 1 and EU - 9 would equal 0 and EU - 1 would equal 0. So U equal 9 and U equal 1. The problem is we weren't trying to solve for U, we were trying to solve for X. So then we would re substitute and we'd get X ^2 equal 9 and X ^2 equal 1. To get rid of the square we would positive and negative square root. So we positive and negative square root and we'd get X equal positive -3 and X equal positive -1. The next example, 90 to the 4th minus fourteen t ^2 + 5. This one, let's see if it can factor. We'd multiply that nine and that five to get 45 S 1 and 45 or 3 and 15 or 5 and 9. Do any of those multiply to give me a positive 45 and add to give me a -14? If it was a -5 and a -9, that would so we could have the factorization T to the fourth would be AT squared times AT squared. And if I put that -5 / 9 and simplify and that -9 / 9 and simplify. Remember, once it's simplified, if there's a denominator, we're going to move it in front to be the coefficient. So nine t ^2 - 5 and t ^2 - 1, each of those parentheses are going to be set equal to 0. But we realize that t ^2 - 1 is actually still factorable because that's t - 1 and t + 1 by difference of squares. So now nine t ^2 - 5 can't factor anymore, so I'm going to let it equal 0. I'm going to have t - 1 = 0 and t + 1 = 0. So 9 T squared would equal 5T squared would equal 5 ninths. When we square root it, we'd get positive negative sqrt 5 / 3, so we'd also have T equal 1 and we'd have T equal -1. So when we list all of our answers, we're going to have -1 negative root 5 / 3, root 5 / 3, and one. We expected 4 answers because it was a four degree equation. If we look at some more examples, this next one, it's actually has a degree of 1. So we will expect one final answer. Now this one's not going to factor, so we're actually going to need to use quadratic formula. This time we're going to have square root of S equaling are opposite of our B, so opposite of -4 plus or minus the square root -4 ^2 -, 4 * A * C, the C being a -1 all over 2 * a. So if we simplify this up, we get 4 ± sqrt 16 + 4 / 2. Well, 16 + 4 is 20, and we could think of that 20 as 4 * 5. We know sqrt 4 is 2, we don't know sqrt 5, so we're going to leave it. The numerator then has a two that can factor out. So I get 2 * 2 plus or minus root 5 / 2 monomial over monomial. So I'm going to get 2 ± sqrt 5. So we have square root of S equaling 2 + sqrt 5. And we also have square root of S equaling 2 -, sqrt 5. Well, square root of S because it's a square root, has to equal a positive #2 plus root 5. Is that a positive number? Well, positive plus a positive is +2 minus root 5 is not a positive number because sqrt 5 is more than sqrt 4. So this one's not going to give us a solution. So the only one we have to think about is the square root of S equal 2 plus root 5. How do I get rid of a square root? We're going to square it. And if I square one side, we're going to have to square the other on that right hand side. We're going to have to remember to foil. So if I have two plus root 5 * 2 plus root 5, we're going to get S equaling 4 +2 root 5 + 2 root 5 + sqrt 5 ^2, or S is going to equal 9 + 4 sqrt 5. On the next example, we're going to do the same thing. This one actually will factor what 2 numbers multiply to give us -4. But add to give us -3 and the answer is going to be a -4 and a +1 sqrt X * sqrt X multiplies to give me the X. So we're going to have sqrt X -, 4 equaling 0 and sqrt X + 1 equaling 0. So square root of X is equaling 4 sqrt X equal -1. A square root always has to equal a positive number, so the square root of X equaling -1's not going to be possible. So over here, when we get rid of the square root by squaring one side, we have to remember to square the other. So we're going to get X = 16 as our final answer. We only expected one solution because the height, the degree of the equation was one. The highest exponent was one. If we look at this one this time we're going to think about this whole 3 + sqrt X * 3 + sqrt X would give us that 3 + sqrt X ^2. Do we have two numbers that can multiply to give us -10 and add to give us three? I'm looking at those coefficients, so +5 times -2 gives me -10 and +5 + -2 gives me a positive three. If we combine our like terms within those parentheses, we get 8 + sqrt X and we get 1 + sqrt X. Now we're going to set each of those parentheses equal to 0. So square root of X equal -8 and that's not going to be possible. Square root of X equal -1 that one's not possible either. So in this example, we're going to have no solution on our next problem. Let's see, can we factor this one? If we multiply the four and the -5 negative 21 and 22 and 10, four and five, we need it to add to give us a +1. So how about a -4 and a +5? So what times? What gives me X to the -2 and it's going to be X to the -1. So we're going to have X to the -1 - 4 / 4 and X to the -1 + 5 / 4. Now, once we simplify that, if it still has a denominator when it's simplified, that denominator is going to go in front of the coefficient. So X to the -1 - 1 and 4X to the -1 + 5 = 0. We're going to set each of these equal to 0. I'm going to add one across. Now what does that -1 really mean? It means 1 / X. So if I cross multiply here, I'm going to get X equaling one for my first possible solution. So I'm going to get 4X to the -1 equal -5 or 4 / X equal -5 four equal -5 X by cross multiplying so X equal negative 4/5. So our two solutions here are negative 4/5 and 1:00. We expect 2 solutions because it was an X to the -2 which was the one over. We actually could have solved this as 4 / X ^2 + 1 / X - 5 equaling 0 and multiplied by a common denominator. If we had done it this way, we would have had 4 + X -, 5 X squared equaling 0. I personally like my X ^2 to be positive so I can multiply this entire thing through by a -1, and if we factored this, we would get the same solutions. So we'd have five X + 4 * X - 1 and it's factored form and X would be negative 4/5 and 1:00 Here we're going to see, can we factor it 2 * -1 is -2 so one and two I would want it to be a -2. So X to the -1 + 1 / 2 and X to the -1 - 2 / 2. When I simplify that, if there's a denominator, we're going to put it in front as the numerator, not as the numerators. The coefficient. So we get 2X to the -1 + 1 = 0 X to the -1 - 1 equaling 0, so 2 / X equal -1 so 2 equal negative X or X equal -2 X to the -1 equal 1. So 1 / X is 1 cross multiply X equal 1. Our two solutions -2 and one this next one. What times? What gives me T to the 2/3? Well with exponents we add them so T to the 1/3 * t to the one third is T to the 2/3. Two numbers that are going to multiply to give me -6 and add to give me a +1 are going to be -2 and +3. So T to the 1/3 -, 2 = 0, T to the 1/3 + 3 = 0, T to the 1/3 equal 2. If we cube each side, we get T to the one third cubed, so those will cancel giving us T and two cubed is 8 T to the 1/3 equal -3. If I cube each side here, we get T equal -27. So our two solutions -27 and 8 here, T to the 1/4 * t to the one fourth would give us T to the half, because once again we have to add the exponents a +1 and a +2 multiply to give us 2 and add to give us 3. So T to the 1/4 + 1 = 0, T to the 1/4 + 2 = 0. Well, the 1/4 is really the 4th root of T equaling a -1. The 4th root. It's an even index, so it has to equal a positive. The fact that it's equaling a -1 tells us that's not possible. The 4th root of T is going to also equal -2, but an even index always has to equal a positive number, so this one is not possible either. So this is a no solution problem. OK, so for the next two examples we want to have all the X intercepts. X intercept just really means that Y is going to equal 0. So when we look at this, are there 2 numbers that multiply to give us -30 but add to give us 13? If the answer's yes, we can factor. If the answer's no, we have to use quadratic formula. So let's see, we need them to multiply to give us -30. So how about a -2? And a +15 would add to give us the 13, so we'd have zero what times what gives us plain X? It's sqrt X * sqrt X so -2 / 5 and sqrt X + 15 / 5. We simplify both those fractions, and if there's a denominator, I bring it in front of the first term. We're going to set each of those parentheses now that it's factored equaling to 0. So 5 square roots of X equal 2. If we divide, we get square root of X equal 2/5. To get rid of a square root, we're going to square it. But if we square one side, we have to square the other, so X is going to equal 420 fifths. This other square root of X equal -3 square root is even index. Even index always has to equal a positive number, so this one's not possible. We only have one solution here, and it is 420 fifths. Now we're actually asked for intercepts. Intercepts are always points, so it's the .420 fifths, 0. When X is 420 fifths, Y is 0, so on this next one we're going to have X ^2 + 1 / X ^2 times itself to give us that 4th power. Are there 2 numbers that multiply to give us 12 and add to give us 4? There aren't 2 numbers that multiply to give us 12 and add to give us 4. So this one we're actually going to have to factor using quadratic formula. So we get X ^2 + 1 / X ^2 equaling the opposite of B plus or minus the square root b ^2 - 4 * a * C. All over 2A. So X ^2 + 1 / X ^2 is going to equal a -4 plus or minus the square root 16 - 48 / 2 X ^2 + 1 / X quantity squared equal -4 ± sqrt -32 / 2. Well, that square root of a negative is an imaginary number, and if it's an imaginary number, we can't have it crossing the real number. Line 32 is going to be 2 * 16, so we're going to get four I sqrt 2 because sqrt 2 we don't know, but sqrt 16 was four and the square root of the negative. So the fact that we have an I here, an imaginary number, means that there are no X intercepts because our graphing is always on the real number line. Thank you and have a wonderful day.