Q: What number must be added to the polynomial to complete the square x2-9x?

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81. To complete the square of x^2 + 18x, you take half the coefficient of the x term (half of 18 is 9), and square that number (9 squared is 81). To confirm this works, you can now factor x^2 + 18x + 81 and see that it factors as (x+9)(x+9), or (x+9)^2, a perfect square.

49-apex

-12

9 needs to be added to complete the square. When squaring (x + n), the result is x2 + 2nx + n2. We have +10x, so n must be 10/2 = 5. Thus: (x + 5)2 = x2 +10x + 25 = x2 +10x + 14 + 9 So 9 needs to be added to complete the square, giving: x2 +10x + 14 = (x + 5)2 - 9

X^2 + X = 0 halve the linear term (1) and square it then add to both sides X^2 + X + 1/4 = 1/4 factor left; gather terms right (X + 1/2)^2 = 1/4 (X + 1/2)^2 - 1/4 = 0 (-1/2,-1/4) vector and the number 1/4 was added to both sides completing the square

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144

-3

49/4 or 12.25

Adding 100 completes the square. { I wonder why you want to complete the square when the expression already factors as x(x-20) }.

81. To complete the square of x^2 + 18x, you take half the coefficient of the x term (half of 18 is 9), and square that number (9 squared is 81). To confirm this works, you can now factor x^2 + 18x + 81 and see that it factors as (x+9)(x+9), or (x+9)^2, a perfect square.

49-apex

-12

A polynomial is several terms added together.

X^2 + X = 0 halve the linear term (1) and square it then add to both sides X^2 + X + 1/4 = 1/4 factor left; gather terms right (X + 1/2)^2 = 1/4 (X + 1/2)^2 - 1/4 = 0 (-1/2,-1/4) vector and the number 1/4 was added to both sides completing the square

Complete the square is the process of creating a "perfect square" polynomial. We call (x + a)^2 a perfect square, where a is a constant. Using simple distributivity of numbers, we get x^2 + 2ax + a^2 is a representation of a perfect square in simplified formed. so (x + a) ^2 = x^2 + 2ax + a^2. Given a degree polynomial in the form x^2 + nx, where m and n are constants, when we "complete the square", we are looking for values that will turn it into something like x^2 + 2ax + a^2. The entire idea is to find what "a" is. 2a is the coefficient for the degree one monomial "2ax" for what we want, also n is the coefficient for the degree one monomial "nx" for what we have. Then why don't we just say n = 2a for some a. To find a, it's obvious a = n/2. We have the degree 2 term (x^2), degree 1 term (nx = 2 . n/2 .x). We need the constant of a^2. a^2 = (n/2)^2 = n^2 / 4. In this case, n = 13.

9

9 needs to be added to complete the square. When squaring (x + n), the result is x2 + 2nx + n2. We have +10x, so n must be 10/2 = 5. Thus: (x + 5)2 = x2 +10x + 25 = x2 +10x + 14 + 9 So 9 needs to be added to complete the square, giving: x2 +10x + 14 = (x + 5)2 - 9