Solve 3x²+6x-7=0 by using completing the square method. Leave your answer in surd form.

This is a question which is simply asking you to solve a quadratic. There are three main ways to do this; quadratic equation, factorising into brackets and completing the square. As the question specifically wants you to use complete the square, you must do so. To use complete the square, the quadratic has to be in the form of ax²+bx+c where a has to equal 1. To make this so, you must factorise the quadratic by the current a value so the new quadratic would be 3[x²+2x] - 7 =0. Now you can use normal completing the square method using the values inside the square bracket. You put them into one bracket by dividing the b value so it now becomes... (x+2/2)² = (x+1)² You then have to subtract the square value of the number inside the bracket. Here it is 1² hence you subtract 1 so the equation now becomes.... 3[(x+1)²-1]-7=0 The reason you have to subtract one is so the original equation is identical to the one we have created. If you expand (x+1)² you get x²+2x+1 which is the same as our new equation barring the +1. Hence to balance this, we must subtract 1. Now it is just expanding and re-arranging. Expanding the bracket gives... 3(x+1)²-3-7=0 Re-arranging gives... 3(x+1) -10 =0 3(x+1)²=10 (x+1)²=10/3 x+1 = ±√ (10/3) x = -1 ±√ (10/3)
Hence our two solutions are: -1+√ (10/3) and -1 -√ (10/3)