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ax 2 + bx + c = 0 If the discriminant is a positive real number, then the associated quadratic has two real roots If the discriminant is equal to 0, then the quadratic has one real root with multiplicity 2 There are plenty of quadratic equations in which the discriminant is a negative real number Here s an example: 4x 2 4x + 36 = 0 This is almost exactly the same equation as we solved in the second example above The only difference is that the second coefficient is 4 rather than 24 The discriminant here is b 2 4ac = 42 4 4 36 = 16 576 = 560 When we apply the quadratic formula, we must take the square root of 560 That s an imaginary number! A negative discriminant gives us imaginary or complex roots We ll explore situations like this in the next chapter We ll also see what happens when one or more of the coefficients in a quadratic equation are imaginary or complex

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Do you wonder how we came up with the constant b 2/(4a 2) to make a perfect square in the process of deriving the quadratic formula Again, this is the sixth sense at work, a form of intuition that you can develop only with practice

0 /4 0

Here s a challenge!

In the derivation of the quadratic formula, we made a quantum leap when we claimed that the polynomial x 2 + (b /a)x + b 2/(4a 2) is a perfect square that can be factored into [x + b /(2a)]2 Show that this is actually true

0 3

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We can work it the other way, multiplying the factors out to get the polynomial Let s rewrite the above squared binomial as a product of sums, like this: [x + b /(2a)] [x + b /(2a)] When we apply the product of sums and product of quotients rules from Chap 9, this becomes x 2 + xb /(2a) + xb /(2a) + b 2/(4a 2 ) We can add the two middle terms together, getting x 2 + 2xb /(2a) + b 2/(4a 2 ) Canceling out the 2 in the numerator and denominator of the middle term, we obtain x 2 + xb /a + b 2/(4a 2) which can also be written as

x 2 + (b /a)x + b 2/(4a 2 )

1 1 0= 2 2

This is an open-book quiz You may (and should) refer to the text as you solve these problems Don t hurry! You ll find worked-out answers in App C The solutions in the appendix may not represent the only way a problem can be figured out If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it! 1 Multiply out the following equation, putting it into polynomial standard form ( 7x 5)( 2x + 9) = 0 2 Factor the following quadratic Then find the roots and state the solution set Here s a hint: The coefficients and constants in the factors are all integers x 2 + 10x + 25 = 0 3 Factor the following quadratic Then find the roots and state the solution set Here s a hint: The coefficients and constants in the factors are all integers 2x 2 + 8x 10 = 0 4 Factor the following quadratic Then find the roots and state the solution set Here s a hint: The coefficients and constants in the factors are all integers 12x 2 + 7x 10 = 0

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