So those are the points where, those are the x-values where the function intersects the x-axis. So, also the point negative one comma zero is on this graph. And also, if x is equal to negative one, negative one minus two, negative three. So, the point five comma zero is going to be on this graph. If x is equal to five,įive minus two is three, squared is nine, minus nine is zero. Well, you add two to both sides of this, you get x is equal to five, or x is equal to, if weĪdd two to both sides of this equation, you'll get Is equal to positive three or x minus two is equal to negative three. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. Sides and so we could get x minus two squared is equal to nine. So we could rewrite this as x, x minus two squared minus nine equals zero. This thing right over here "equal zero?" So, let me just write that down. X-values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying, "For what x-values doesį of x equal zero?" So we could just say, "For what x-values does And that means that ourįunction is equal to zero. Notice our y-coordinateĪt either of those points are going to be equal to zero. Well, in order to intersect the x-axis, y must be equal to zero. The x-values where you intersect, where you intersect the x-axis. It can also utilize other methods helpful to solving quadratic equations, such as completing the square, factoring and graphing. In doing so, WolframAlpha finds both the real and complex roots of these equations. Let's say that the y isĮqual to some other function, not necessarily this f of x. WolframAlpha can apply the quadratic formula to solve equations coercible into the form ax2 +bx+c 0 a x 2 + b x + c 0. If I have the graph of some function that looks something like that. Talking about some graph, so I'm not necessarily gonnaĭraw that y equals f of x. When solving quadratic equations by completing the square, be careful to add to both sides of the equation to maintain equality. And then we're asked at what x-values does the graph of y equalsį of x intersect the x-axis. So, we are told that f of x isĮqual to x minus two squared minus nine. That's presented to us in a slightly different way. So, these are the two possible x-values that satisfy the equation. And when x is equal to negative five, negative five plus three is negative two, squared is positive four, minusįour is also equal to zero. You substitute it back in if you substitute x equals negative one, then x plus three is equal to two, two-squared is four, minus four is zero. Substitute it back in, and then you can see when So, those are the two possible solutions and you can verify that. Negative two minus three is negative five. Or, over here we could subtract three from both sides to solve for x. Sides to solve for x and we're left with x isĮqual to negative one. So, if x plus three isĮqual to two, we could just subtract three from both If x plus three was negative two, negative two-squared is equal to four. Notice, if x plus three was positive two, two-squared is equal to four. And so we could write that x plus three couldīe equal to positive two or x plus three could beĮqual to negative two. Positive square root of four or the negative square root of four. Something right over here, is going to be equal to the If something-squared is equal to four, that means that the something, that means that this Three is going to be equal to the plus or minus When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the x2 x. So, one way of thinking about it is, I'm saying that x plus Way of thinking about it, if I have something-squared equaling four, I could say that that something needs to either be positive or negative two. This also applies to radicals with other even indices, like 4th roots, 6th roots, etc. So yes, you are correct that a radical equation with the square root of an unknown equal to a negative number will produce no solution. And so now, I could take the square root of both sides and, or, another In general, when we solve radical equations, we often look for real solutions to the equations. So, x plus three squared is equal to four. So, adding four to both sides will get rid of thisįour, subtracting four, this negative four on the left-hand side. This is I'm gonna isolate the x plus three squared on one side and the best way to do that \).The video and see if you can solve for x here.
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