Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Quadratic like equations. Certain equations that are not really quadratic can be thought of as quadratic like because they behave like a quadratic. We could think about letting U equal X ^2. If U is X ^2, then U squared would be X to the 4th. Now this would turn in to a quadratic equation, and we could think about what 2 numbers multiply to give us 9 but add to give us -10 and U -, 1 U -9. But you remember it's really X ^2. So this is really X ^2 -, 1 X squared -9. And we could see that those are really 2 differences of squares. So we would get for our final four answers because it was a fourth degree. So we expect 4 answers X equaling positive -1 positive -3 for this next one. If we wanted to think about using our U again, we'd let U = sqrt X and U square would equal X, so we'd have U ^2 -, 3, U -4 = 0, so 2 numbers that multiply to give us -4. But add to give us 3 would be -4 and one. So U equal 4 and U equal -1. But U is square root of X, so square root of X is 4 and square root of X is -1. We square root side. We get X = 16 if we square each side here we know the square root is never going to equal a negative number, so this piece isn't possible. So we get X = 16. And we'd actually only expect one answer because the highest degree here was exponent of one. This next one we're going to let U equal T to the one third, so U squared is T to the 2/3. So we'd get U ^2 + U - 6 = 0, so U + 3 and U - 2, the two numbers that multiply to give us -6, but add to give us 1. So U is -3 and U is 2. But we're not trying to solve for U. So T to the one third is -3 T to the one third is 2. How do we get rid of the one third? We take it to the third power, and if we do one side, we do the other. So we get T to the -3 ^3 and T to the two cubed. So T is going to be -27 and +8. Now the fractional exponents don't work the same as whole number exponents. We might guess that there's two because it was 2/3, but that won't always be the case. So on this next one, we would let U equal three plus square roots of X. So U squared would then be 3 plus square roots of X ^2. So we get U ^2 + 3, U -10 = 0, U +5 U -2. The two numbers that multiply to give us -10 But I have to give us 3. So U equal -5 and U equal 2. Now we're not solving for you, we're solving for X. So we have 3 + sqrt X equal -5 and 3 + sqrt X equal 2. If we subtract, we get square root of X equaling -8. That's not possible because square root has to be equaling a positive or zero. Here's square root of X equaling -1. That's not possible. So this one actually has no solution. So it's final answer is no solution. We have a few for you to try. There's a couple on this page, and then the next page actually starts with a couple for you to try. Also, the next type is to find all X intercepts. And when we're talking about an X intercept on a graph, we're really just saying the Y is 0. And remember, F of X really acts like our Y. So in this case, we want 0 equaling five X + 13 square roots of X -, 6. If I let U = sqrt X, then U squared equal X so we get 0 equaling five U ^2 + 13 U -6. So when we solve this one, we're going to think about two numbers that multiply to give us -30 but add to give us 13, so 132 and 15, that one -2 positive 15. So I'm going to have U -, 2 fifths, U + 15 fifths. So 0 equal 5 U -2 * U + 3. Once again, we're not solving for U, we're solving for X. So we'd get U equal 2/5, but U was square root of X. So when we square that, we get X equal 425th and we get U equaling -3 sqrt X equal -3. That one's not possible. So we only get the one answer of four 25th. But it's an intercept, so we're actually going to give it as a point. It's the point 425th 0 on a graph. This next one let U equal X ^2 plus 1 / X quantity squared so that our U ^2 is this 1. So 0 equal U ^2 + 4 U plus 12/2 numbers that multiply to give me 12 and add to give me four. Well, this isn't going to factor easily, so we're actually going to have to use quadratic formula. So U equal the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over 2A so -4 ± sqrt -32. The fact that this is going to be an imaginary number tells me we don't have a graph. We won't have any X intercepts because the graph is over the real numbers. So what this really says is we're going to get a graph that's never going to cross the X axis. So there are no X intercepts for this one. And then we have a couple for you to try if we come back here. Oh, I skipped one way back here. Sorry. I was like, I wanted to make sure you guys got an example of this other type. So when we look at this one, let's let U equal X to the -1. So U ^2 = X to the -2. So we get four U ^2 + U - 5, equaling 02 numbers that multiply to give us -20, but add to give us one would be 4 and 5 S 4 UU plus 5 - 1. So our U is -5 fourths here, and our U is 1 here. So we're not solving for U, we're solving for X. So we have X to the -1 equal -5 fourths and X to the -1 equal 1. Well, remember the negative exponent really means one over. So in order to solve this, we could just think about flipping both sides of the fraction. So here we get X equal -4 fifths and here we get X equal 1 / 1 which is just one. Thank you and have a wonderful day.