Write x^2 + 6x - 10 in the form ((x+a)^2)+b?
To find the value of a divide the coeffecient of x. Here this would be 6/2 which = 3. If we were to expand (x + 3)^2 + b this would give us:x^2 + 3x +3x + 3^2 +b which simplifies to x^2 + 6x + 3^2 + b. If we compare the coeffecients of this with the given quadratic, we can see the constant term is 3^2 + b = - 10If we rearrange this we can see b = -19Therefore x^2 + 6x - 10 in the form ((x+a)^2)+b is (x + 3)^2 - 19.The b is always equal to a^2 - [constant term of the quadratic].
GG
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