![]() The x-values where you intersect, where you intersect the x-axis. Let's say that the y isĮqual to some other function, not necessarily this f of x. 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. 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 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. And so now, I could take the square root of both sides and, or, another 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 10.The video and see if you can solve for x here. Find the solution of quadratic equation 6x 2-13 23 using square roots.Find the solution of quadratic equation 5x 2 – 1 = 24 using square roots.Find the solution of quadratic equation x² + 1 = 1 using square roots.Find the solution of quadratic equation x 2 – 1 = 1 using square roots.Find the solutions of the equation x 2 = 81.Find the solutions of the equation x² = 9.Find the solutions of the equation x 2 = 16.Find the solutions of the equation x² = 45.Find the solutions of the equation x 2 = 1.The length of the ladder =√365 ≈ 19.1 m Exercise X 2 = 13²+14², from the Pythagorean theorem.Īs the length of the ladder cannot be negative The solutions of the quadratic equation are x = +√25 and x = – 25 Real Life ExampleĪ ladder is leaned on a wall, the height on the wall is 13 m, the ladder is 14 m away from the wall, what is the length of the ladder? Step1: Given quadratic equation 3x 2+9 = 69.… (1) The solutions of the quadratic equation are x = +√10 and x = −√10įind the solution of quadratic equation 3x 2+9 = 69 using square roots. Step1: Given quadratic equation 3x 2−4 = 26 … (1) The solutions of the quadratic equation are x = +5 and x = – 5įind the solution of quadratic equation 3x 2−4 = 26 using square roots. Step1: Given quadratic equation x 2– 1 = 24 … (1) The solutions of the quadratic equation are x = +4 and x = -4įind the solution of quadratic equation x 2– 1= 24 using square roots. Step1: Given quadratic equation 4x 2 +5 = 69 … (1) Take the square root on each side of the equation.įind the solution of quadratic equation 4x 2+5 = 69 using square roots. ![]() How to solve an equation in the form of ax 2+b=c?įirst write the equation in the form of x 2=a, where a is a real number. Solve Quadratic Equations of the form □□ □+□=□ There is no real number that can be multiplied to get a negative number for which square root can be obtained. The solutions of the quadratic equation are x = +8 and x = -8.įind the solutions of the equation x² = -36. Step2: By seeing the equation we remember that 64 is square of 8. The solutions of the quadratic equation are x = +12 and x = -12.įind the solutions of the equation x² = 64. Step2: By seeing the equation we remember that 144 is square of 12. The solutions of the quadratic equation are x = +10 and x = -10.įind the solutions of the equation x 2 = 144. Step2: By seeing the equation we remember that 100 is the square of 10. ![]() The solutions of the quadratic equation are x = +11 and x = -11.įind the solutions of the equation x 2 = 100.
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