Answer:
(c) x^2+4x-21=0
(b) (x-y)(x+y)
Explanation:
1.
A quadratic is conveniently solved using the technique of completing the square when the leading coefficient can be factored from the variable terms, and the resulting linear term has an even coefficient.
2x^3-12x^2+5x-30=0
Not a quadratic. When the only real root is factored out, the result is ...
(x -6)(2x^2 +5) = 0
The quadratic factor has no linear term, so "completing the square" cannot be used.
5x^2+7x=12
The leading coefficient is not a factor of the linear term, so using "completing the square" gets messy. That may not be the best solution method for this.
5x^2 +7x = 12 ⇒ 5(x^2 +1.4x) = 12 ⇒ 5(x +0.7)^2 = 12 +5(0.49)
⇒ x = -0.7 ± √(14.45/5) = -0.7 ±1.7
Solution is possible, but fractions get involved right away. Completing the square may not be the best choice of solution method.
x^2+4x-21=0
The leading coefficient is 1, and the linear term has an even coefficient. These are the ideal conditions for solution using "completing the square."
x^2 +4x -21 = 0 ⇒ (x +2)^2 = 25 ⇒ x = -2 ±5
x^2=81
There is no linear term, so "completing the square" cannot be used.
x^2+4x-21=0 is best solved using completing the square
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2.
We can see which of these patterns would be useful for factoring the difference of squares by expanding each of them.
(x-y)^2 = x^2 -2xy +y^2 . . . . not the difference of squares
(x-y) (x+y) = x^2 -xy +xy -y^2 = x^2 -y^2 . . . . the difference of squares
(x+y) (x+y) = x^2 +2xy +y^2 . . . . not the difference of squares
2(x-y) = 2x -2y . . . . not the difference of squares