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If you know one root of a given quadratic equation, how can you calculate the other: (1) from the coefficient of x, (2) from the absolute term, (3) by the Factor Theorem? each of these methods in turn, to find the second root in each of the following equations:

13. x2 -8.7 x + 17.6 = 0, one root 5.5.

14. x2 - 2.1 x + .9 = 0, one root 1.5.

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=

0, one root - 5.1.

16. x2+11.8 x + 34.17

17. x2+3.6 x — 25.92 = 0, one root - 7.2.

18. x2 - .085 x + .00175 = 0, one root .05.

NATURE OF THE ROOTS

136. We shall discuss the nature of the roots of the quadratic equation ax2 + bx + c = 0, in which the letters a, b, c are assumed to be real and rational numbers. When some of these letters are imaginary or irrational, the conclusions which we shall draw do not necessarily hold. ax2 + bx + c = 0 is

The solution of

- b ± √b2.

4 ac

x=

2 a

The nature of the roots depends upon b2 called the discriminant.

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I. When b2-4 ac=0, the roots are real and equal. Explain. II. When b2. 4 ac > 0, the roots are real and unequal. If b2-4ac is a perfect square, both roots are rational; if

b2 - 4 ac is not a perfect square, both roots are irrational. Why?

III. When b2 4 ac< 0, both roots are imaginary. Explain.

To illustrate: 1. In 2

10x+25=0, b2-4ac = 100 — 100 = 0.

Hence the roots are real, rational, and equal. The roots are 5, 5.

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3. In x2 - 2x + 6 = 0, b2 — 4 ac = 424 = =-20. Hence both roots are imaginary.

EXERCISES

137. Determine, without solving, the nature of the roots of:

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For what values of a are the roots of the following equations equal? Real and unequal? Imaginary?

11. ax2+4x=— - 1.

12. x2+ ax + 5 = 0.

13. 3x2 + 3x + a = 0.

14. ax2 + ax + 2 = 0.

GRAPH OF THE QUADRATIC EQUATION, ax2 + bx + c = y 138. The value of the expression, ax2 + bx + c, for given values of a, b, and c, depends upon the value of the variable x. For that reason the expression is called a function of x. If we write ax2 + bx + c = y, then for every value of x there is a corresponding value of y.

The values of x and the corresponding values of y may be taken as the coördinates of points on the graph of the equation ax2 + bx + c = y.

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Thus it will be seen (Fig. 17) that the graph crosses the x-axis at the points (5, 0) and (-1, 0). Upon solving x2 - 4x-5

= 0, we +5 and 1, and these are the

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find that x =
values of x which make y = 0.

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