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91. Solution of quadratic equations by factoring. When the left-hand member of an equation can be factored readily, this is the most convenient method of solution. It also illustrates very clearly the meaning and property of the roots of the equation.

EXAMPLE. Solve

x2+2x-15= 0.

Factor the left-hand member, (x+5) (x − 3) = 0.

The object in solving an equation is to find numbers that substituted for the variable satisfy the equation. But since zero multiplied by any number is zero (p. 3), any value of x which causes one factor of an expression to vanish makes the whole expression vanish. If in this case x = 3, our equation in factored form becomes

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Thus the numbers 3 and - 5 are solutions of the equation.

92. Solution of an equation by factoring. We have immediately the

RULE. Transpose all the terms to the left-hand member of the equation.

Factor that member into linear factors.

The values of the variable that make the factors vanish are roots of the equation.

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5. 13x238 x = 3. 7. x2 - 40x+111 0. 9. x2 - 18x208 = 0. 11. x2 - 3 ax - 4a2 = 0.

13. (x2 - 1) + (x − 1)2 = 0.

2. 6x2+7x= 3.

4. 5x217 x + 6 = 0.
6. 2x2 - 5x – 25 = 0.
8. 13x240 x + 3 = 0.
10. 3x226x + 35 =
= 0.

12. (x − a)2 - (x − b)2 = 0.
14. (3x-5)2 - (9x + 1)2 = 0.
15. (2 x − 1)(x + 2) + (x − 1) (x − 2). — — 4.
16. (7x1)(x+3)-(4x-3)(x-1)=24.

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2(x+2).

x2

x − 4 = 4√x – 7.

8x1616x

x224 x + 128 = 0.

(x - 16) (x — 8) = 0.

x=16, x = 8.

- 112.

Check: 316 4 √16 – 7 − 2 (16 + 2) = 48 − 12 – 32 − 4 = 0.

3.8

·4 √8 −7 – 2 (8 + 2) = 24 − 4 — 20 = 0.

In the following examples, as always, the quadratic equation should be solved by factoring when possible. Recourse to the longer but sure method of completing the square is always available.

When an equation is cleared of fractions or squared in the process of bringing it into quadratic form (1), § 88, extraneous solutions may be introduced. The results should be verified in every case and extraneous solutions rejected.

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48. √(a + x) (x + b) + √(a − x) (x − b) = 2 √ax.

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49. 2√2a + b + 2 x + √10 a + b 6 x 10 a 9b − 6x.

93. Quadratic form. Any equation is in quadratic form if it may be written as a trinomial consisting of a constant term and two terms involving the variable (or an expression which may be considered as the variable), the exponent in one term being twice that in the other. By the constant term is meant the term not containing the variable.

Thus x- · 8 √x + 13 = 0, x ̄ § + x ̄3 − 3 = 0, a2x−2n. 22-2x- 3 √x2-2x

(a + b) x−n + b2 = 0, 3 +17 0 are all in quadratic form. In the last the

whole expression x22x3 is taken as the variable.

It is usually convenient to replace by a single letter the lower power of the variable or expression with respect to which the equation is in quadratic form.

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Solution: Add - 3 to both members and rearrange terms,

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Let

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35. (2x2-3x+1)2=22x2-33x+1. 36. (x2-5x+7)2 - (x-2) (x-3)=1. 37. x2+5=8x+2√x2-8x+40. 38. 2 √(x − 4)2 + 4 (x − 4)-3 – 9.

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94. Problems solvable by quadratic equations. The principle of translating the problem into algebraic symbols, explained in § 55, should be observed here. The result should be verified in every case. It may happen that the problem implies restrictions that are not expressed in the equation to which the problem leads. In this case some of the solutions of the equation may not be consistent with the problem; for instance, when the

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