How do I solve quadratics by completing the squares? I’ve looked it up, but I don’t get it and I have homework due on it?
Hello Isabella, Quadratic Equation is of the form ax^2 + bx + c where x is a variable of degree 2 and a, b and c are constants. For example, x^2 + 5x + 6 Simple rules, Step 1: multiply constant ‘a’ with constant ‘c’ for solving the middle term (bx) and breaking it down into two components. So, here a = 1 and c = 6 Thus, a*c = 1*6 = 6 Step 2 : Now we need to break middle term 5x in the above example such that addition of two components resulting 5x should give me output 6 when we multiply the two components. Here 5x = 2x + 3x Thus, our first component : ‘2’ Our second component : ‘3’ Also we see that 2*3 = 6 which is multiplication of two components resulting 6 So, we rewrite our quadratic equation as, x^2 + 5x + 6 = x^2 + 2x + 3x + 6 = x (x + 2) + 3 (x + 2) = (x + 3) (x + 2) Hope this helps you, for better understanding just keep our steps of simple rules and you will enjoy learning quadratic equations.
Halve it, Square it, Take it away. (focus on the number next to the x) Using an example x^2+10x+6 firstly you halve the 10 => 5 Then you put this in a bracket with the x and square the bracket => (x+5)^2 Then you square it, square the 5 =>25 Finally, you take it away => (x+5)^2 -25 +6 Then you have to tidy everything up => (x+5)^2 - 19 As a little explanation as to why. You halve the 10 because when you expand the brackets you will end up with x^2 +5x + 5x + 25 which simplifies to x^2 + 10x + 25. If you did not halve it you would end up doubling the amount of x's, in this case ending up with 20x You square and take away the 5 because when you expand the brackets at the end, you should now be left with the original equation. If you did not square it and take it away, you would be left with an excess 25 at the end. Hope this helps. :)
Best to give an example x^2 -6x +8=0 can factorise it easily but by completing the square you do this (x-3)^2 +8 -9 =0 so (x-3)^2 - 1= 0 So basically you half the number of the middle term with x you put in a bracket so when you square it you get the +9 so you need to subtract this then you shift to RHS and solve: (x-3)^2 - 1 = 0 Therefore (x-3)^2 = 1 So (x-3) = + or - Sqr(1) Gives x = 3 + or - 1 x = 3 + 1= 4 Or x = 3 - 1 = 2 of course in this case much easier to factorise and get solution as a check x^2 -6x +8=0 this factorises to (x-4)(x- 2)=0 Giving x= 4 or x=2
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