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Such a root is said to have multiplicity 2 If the above polynomial expressions are placed on the left sides of equations with 0 on the right, we get these quadratics: x 2 + 2x + 1 = 0 x 2 2x + 1 = 0 x 2 + 4x + 4 = 0 9x 2 + 12x + 4 = 0 Respectively, they factor into: (x + 1)2 = 0 (x 1)2 = 0 (x + 2)2 = 0 (3x + 2)2 = 0 Because the expressions on the left sides of the above equations are equal to 0, we can take the square roots of both sides in each case without plus-or-minus ambiguity This gives us the following first-degree equations, respectively: x+1=0 x 1=0 x+2=0 3x + 2 = 0 The solutions to these first-degree equations are easy to find: x = 1 x=1 x = 2 x = 2/3 These are the roots of the original quadratics We can write down the solution sets as the single-element sets { 1}, {1}, { 2}, and { 2/3}

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Positive numbers on the right Now we ll depart slightly from the polynomial standard form Let s use the same perfect squares as above, but set the right sides to positive numbers rather than 0 Here are the new equations:

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The left sides can be factored exactly as before, getting (x + 1)2 = 1 (x 1)2 = 4 (x + 2)2 = 16 (3x + 2)2 = 25 If we take the square roots of both sides in each case here, remembering to include both the positive and negative results, we obtain x + 1 = (1)1/2 x 1 = (4)1/2 x + 2 = (16)1/2 3x + 2 = (25)1/2 These simplify to the following double-barreled first-degree equations: x + 1 = 1 x 1 = 2 x + 2 = 4 3x + 2 = 5 Let s think of these as equation pairs: x + 1 = 1 or x + 1 = 1 x 1 = 2 or x 1 = 2 x + 2 = 4 or x + 2 = 4 3x + 2 = 5 or 3x + 2 = 5 When we solve these four pairs of first-degree equations, we get the following results, which are the roots of the original quadratics: x = 0 or x = 2 x = 3 or x = 1 x = 2 or x = 6 x = 1 or x = 7/3 We can write the solution sets respectively as {0, 2}, {3, 1}, {2, 6}, and {1, 7/3} I ll let you plug these values into the original quadratics to be sure we ve found the right roots! Here are the original quadratics once again, in order: x 2 + 2x + 1 = 1 x 2 2x + 1 = 4

g(x) = [x 2 4]/[x 2]

Morphing into a perfect square When we come across a quadratic equation in polynomial standard form, we can sometimes add a positive real number to both sides, getting a perfect square on the left side That will leave us with a nonzero value on the right, but as long as the left side is a perfect square, we can solve the equation just as we did in the four cases above We can take the first quadratic from the above list and subtract 1 from each side, getting

x 2 + 2x = 0 We can take the second equation and subtract 4 from each side, getting x 2 2x 3 = 0 In the third equation, we can subtract 16 from each side to obtain x 2 + 4x 12 = 0 Finally, in the fourth equation, we can take 25 away from each side and get 9x 2 + 12x 21 = 0 If you add the right constants to both sides of each of these equations, you ll get quadratics with perfect squares on their left-hand sides

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outside of the rectangle and part of it is the outside of the oval Any element inside the region bounded by the dashed line is counted only once

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