And so this is really interesting, we have something squared is equal to two times that something. That's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. Have two X minus three squared on the left-hand side, on the right-hand side Or a simpler way to do this if you really payĪttention to the structure of both sides of this equation. Of a classic quadratic form, but there might be a faster And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. Try to find the solutions to this equation right over here. I have ended up with 3 binomials two of which is what Sal got, (2x-5) and (2x-3) and an additional one: (2x+3) Use grouping, extracting a factor from both sides Multiply coefficient on 1st term with that of the last term to get a number where to look for sum of factorsĦ*10=60 and 6 -10= -4 <- Error, both of them need to be negative if I want their product to be a positive 60, this condition seems impossible, didn't notice at the time, counter-intuitively I still got both of Sal's results but also an additional one.Ĭan now split the middle term with my erroneous factors Get all of them on one side (-4x and +6 to both sides) I have several questions relating to the way I tried to solve it (which gave me a really counter-intuitive result):ġ) are there any factors of a (positive) 60 whose sum would be -4? (the 60 is positive, the 4 is negative, specifically)Ģ) I suspect the answer to previous is "no"? Why is that? I thought a trinomial can always be factored using sum-product method?ģ) also did I expand these parentheses correctly (2x -3)^2 -> 4x^2 +9 The answer can also be written as, if rationalized.ETA: I found the mistake - the very first step of expanding the parentheses, after doing it correctly everything worked perfectly using my usual method :D Solve for x: Don't forget that you must include a ± sign when square rooting both sides of any equation. Add that value to both sides of the equation: Half of the x‐term's coefficient squared. Move the constant so it alone is on the right side:ĭivide everything by the leading coefficient, since it's not 1: Take the square roots of both sides of the equation, remembering to add the “±” symbol on the right side.Įxample 3: Solve the quadratic equation by completing the square. Write the left side of the equation as a perfect square.ĥ. Add the constant value to both sides of the equation.Ĥ. If a ≠ 1, divide the entire equation by a.ģ. In other words, move only the constant term to the right side of the equation.Ģ. The most complicated, though itself not very difficult, technique for solving quadratic equations works by forcibly creating a trinomial that's a perfect square (hence the name). Note that the quadratic formula technique can easily find irrational and imaginary roots, unlike the factoring method. You can also write the answers as, the result of multiplying the numerators and denominators of both by −1. The coefficients for the quadratic formula are a = −4, b = 6, and c = −1: You should memorize the quadratic formula if you haven't done so already. A word of warning: Make sure that the quadratic equation you are trying to solve is set equal to 0 before plugging the quadratic equation's coefficients a, b, and c into the formula. This method is especially useful if the quadratic equation is not factorable. If an equation can be written in the form ax 2 + bx + c = 0, then the solutions to that equation can be found using the quadratic formula: Plug each answer into the original equation to ensure that it makes the equation true.Īdd 13 x 2and −10 x to both sides of the equation:įactor the polynomial, set each factor equal to 0, and solve.īecause all three of these x‐values make the quadratic equation true, they are all solutions. Set each factor equal to 0 and solve the smaller equations.Ĥ. Move all non‐zero terms to the left side of the equation, effectively setting the polynomial equal to 0.ģ. To solve a quadratic equation by factoring, follow these steps:ġ. Of those two, the quadratic formula is the easier, but you should still learn how to complete the square. The other two methods, the quadratic formula and completing the square, will both work flawlessly every time, for every quadratic equation. The easiest, factoring, will work only if all solutions are rational. There are three major techniques for solving quadratic equations (equations formed by polynomials of degree 2).
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