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Now that we ve taken four solutions and manufactured four problems from them, let s retrace our steps and get the solutions back In this way, we can get a good feel for how completing the square actually works Imagine that we re confronted with the following four quadratics in polynomial standard form: x 2 + 2x = 0 x 2x 3 = 0
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x 2 + 4x 12 = 0 9x 2 + 12x 21 = 0 We can take the first of these equations and add 1 to each side, getting x 2 + 2x + 1 = 1 That gives us a perfect square on the left side (Recognizing perfect squares when they appear in polynomial form is a sixth sense that evolves over time, and it takes practice to develop it) Factoring, we obtain (x + 1)2 = 1
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The Quadratic Formula
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We can take the square root of both sides and get x + 1 = 1 which can be expressed as the pair x + 1 = 1 or x + 1 = 1
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The solutions are found to be x = 0 or x = 2, so the solution set is {0, 2} The other three equations can be worked out in similar fashion
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Do you wonder what happens if, in order to complete the square in a quadratic, you must subtract a positive number from both sides, getting a negative number on the right side That s a good question In that case, the roots turn out to be imaginary or complex We ll deal with such equations in Chap 23
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Go through maneuvers similar to those we just completed, but with the second, third, and fourth quadratics from above: x 2 2x 3 = 0 x 2 + 4x 12 = 0 9x 2 + 12x 21 = 0
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Solution
You re on your own! Start with perfect squares on the left sides of the equals signs and positive numbers on the right, and then take away those positive numbers from both sides to unsquare the equations
The Quadratic Formula
The technique of completing the square can be applied to the general polynomial standard form of a quadratic equation This gives us a tool for solving quadratics by brute force : the so-called quadratic formula
Deriving the formula Remember the polynomial standard form where x is the variable, and a, b, and c are realnumber constants with a 0 The general formula is
ax 2 + bx + c = 0 Let s rewrite this as ax 2 + bx = c
376 Quadratic Equations with Real Roots
Because a 0 in any quadratic equation, we can divide each side by a, getting x 2 + (b /a)x = c /a It s tempting to think that there must be some constant that we can add to both sides of this equation to get a perfect square on the left side of the equals sign It takes some searching, but that constant does exist It is b 2/(4a 2) When we add it to both sides of the above equation, we obtain x 2 + (b /a)x + b 2/(4a 2) = c /a + b 2/(4a 2) We can now factor the left side into the square of a binomial to get [x + b /(2a)]2 = c /a + b 2/(4a 2) The two terms in the right side of this equation can be added using the sum of quotients rule from Chap 9 to obtain [x + b /(2a)]2 = ( 4a 2c + ab 2) / (4a 3) Canceling out the extra factors of a in the numerator and denominator on the right side, we get [x + b /(2a)]2 = ( 4ac + b 2) / (4a 2) Let s rewrite the numerator on the right side as a difference, so the equation becomes [x + b /(2a)]2 = (b 2 4ac) / (4a 2) If we take the square root of both sides here, remembering the negative as well as the positive, we get x + b /(2a) = [(b 2 4ac) / (4a 2)]1/2 The denominator in the right side is a perfect square; it s equal to (2a)2 Therefore, we can simplify the expression on that side of the equals sign a little bit, considering it as a ratio of square roots rather than the square root of a ratio We obtain x + b /(2a) = (b 2 4ac)1/2 / (2a) If we subtract b /(2a) from both sides, we get x = (b 2 4ac)1/2 / (2a) b /(2a) which expresses x in terms of the constants a, b, and c (finally!) An equation that states the general solution to an unknown is called a formula
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