Solve
replace ∗ with a monomial so that the derived equality will be an identity:
1. (3x + ∗)^2 = 9x^2+6ax+a^2
2.(∗−9c)^2 = 36a^4 - 108a^2c + 81c^2

Step-by-step explanation:

When you square a binomial, you get a result that looks like ...

  (ax +b)^2 = a^2·x^2 + 2abx + b^2

By matching the terms on the right, you can figure out what the terms on the left need to be.

1. The constant term of the expanded form is the square of the constant in parentheses. That constant is a^2, so the value that replaces * in parentheses is a.

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2. The leftmost term in the expanded form is the square (6a^2)^2. That is the square of the * term in the binomial, so the value that replaces * in parentheses is 6a^2.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i