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Hi, I'm a year 12 student and I'm currently doing specialist mathematics. To do this question you need to know a little trick, ill walk you through it. If you have trouble following my working out try going slow and get out a pen and paper ;)

With ax^2+bx+c, you first multiply a and c, and keep that number in your mind, in this case its 10*-32=-320. now look at b, and try to find two numbers, which when multiplied equal -320 and when added they equal -4. I figured the best two numbers are -20 and 16. Now, you split the b term, as -20x+16x=-4x, you can substitute that in for -4x; 10x^2-20x+16x-32x=0. Now that that's done you find common factors (10x and 16) in the equation and split it up in half like this: 10x(x-2)+16(x-2). Next you put the common factor (x-2) at the front of the equation like so, (x-2)(10x+16). so x=2 & x=-16/10=-8/5. I hope I helped! :D:D

Completing The Square Practice Problems

10x² − 4x – 32 = 0

or, 2(5x² ‒ 2x ‒16) = 0 ,take 2 as common

or,5x² ‒ 2x -16 =0 ,compare with Standard Form ax²+bx+c=0

or, 5x² ‒ 2x =16 ,take value without variable to rhs

or, x² ‒ (2/5)x = 16/5 ,divide both sides by coefficient of x²

or, x²-2.x(2/10)+(2/10)²=(16/5)+(2/10)², add sq of the half of coefficient of the current equation to both sides

or,(x-(2/10))²=324/100 ,express lhs as a complete square

or, (x-(2/10))=+18/10,-18/10

Therefore, x=+18/10+2/10=2, x=-18/10+2/10=-8/5

Hope this will help you in solving other such questions in a methodical way.