to the curve. If you should raise the curve in example 1 (move it in a positive direction upwards), the points H and E would approach each other until they would fall together at x=. Should you raise the curve entirely above the X-axis, the intersection would be imaginary (would not exist) and the roots would be imaginary. Let us examine a case of this kind. Here the curve does not cross the X-axis. The intersection of the curve and the X-axis is imaginary (c). This gives us a new kind of number, the square root of a negative number. We have found that the squares of all real numbers are positive, but the square of this number is negative. We have, therefore, found a new unit, and it is called an imaginary number. Any even root of a negative number is imaginary. This new unit is such that its square is negative. (√— 2)2 = (√?√— 1)2 = [2a • (− 1)*]2 = 2(− 1) = — 2. These imaginary numbers often occur in the solution of quadratic equations. EXERCISE 90 Plot the graphs of the following equations. In each case solve the equation algebraically and compare solutions with results in your graph. 236. It is possible to tell the nature of the roots of a quadratic without solving the equation. When we solved example 12, exercise 89, we found, c is the absolute term in the first member, and the second member is zero. In roots (1) and (2) the expression √2-4 ac occurs, the difference in these roots being that in root (1) the radical is added, in root (2) it is subtracted. (a) If b2-4 ac is positive, that is, b2 — 4 ac > 0, then b2 > 4 ac, and there is a real number to add to the -b in the numerator. The roots are then real and unequal. (b) If b2 - 4 ac = 0, that is, if b2 = 4 ac, there is nothing to add to the b, the roots are each real and equal. b 2 a ; namely, the roots are (c) If b2-4ac < 0, that is, if b<4 ac, the quantity under the radical sign is negative, and the roots imaginary. Imaginary roots always go in pairs. Try this test on: Example 2, exercise 90, a =1, b=1, c=— 6, b2 −4 ac = 1+ 24 = +, (real and unequal). Example 9, exercise 90, a = 1, b=4, c=4, b2-4 ac 16-160, (equal). = Example 10, exercise 90, a = 1, b = − 1, c = + 1, Find the nature of the roots of the following: 1. 4x+4x+1=0. 2. 4 x2+8x+1 = 0. 3. 4-12x+9=0. 4. 4x-12x+12=0. 5. 2x2-8x=9. 6. 7 x2 - 4 x − 8 = 0. 7. 7 x2+4x+8=0. 9. 3+14=16. 10. 5x-7x+14=0. 11. Solve examples 1 to 10, using the roots of ax2 + bx+c=0 (§ 236) as a formula. E.g. Solve 5 x2 — 7 x = 2. Reducing to the form ax2 + bx + c = 0, CHAPTER XV The Circle 237. We learned in § 62 that the circle is a plane figure bounded by a curve all points of which are equally distant from a point within called the center. A central angle is an angle whose vertex is at the center of a circle and whose sides are radii of the circle. We will assume that the number of degrees in the angle is equal to the number of degrees in the arc. That is, number B Ө of degrees in ≤ 0= number of degrees in arc AB. A quadrant is one fourth of a circumference. A 238. We can now state the proposition, that in the same circle or in equal circles, equal central angles intercept equal arcs. In the equal circles whose centers are at O and O', respectively, show by superposition that AB = A'B'. 239. It follows from § 238 that in the same circle or in equal circles the greater central angle intercepts the greater arc. 240. Show that a diameter bisects the circle and the circum D M о C ference. 241. A chord is a straight line joining any two points of a circumference, as CD. A segment is the portion of a circle between a chord and its arc, as CDM. The arc CMD is subtended by the chord CD. Every chord subtends two arcs. Illustrate. 242. A straight line cannot meet a circumference in more than two points. 243. An inscribed angle is an angle whose vertex is on the circumference and whose sides are chords; as ABC. The angle ABC is also said to be inscribed in the segment ABC. The angle A is inscribed in segment CAB. THEOREM LVIII B е C 244. In the same circle or in equal circles, equal arcs are subtended by equal chords. Use figure in § 238. Draw chords BA and B'A'. Prove Δ ΟΑΒ=Δ Ο' Α' Β'. 245. The converse of this theorem is also true. State the converse. THEOREM LIX 246. In the same circle or in equal circles, the greater of two arcs is subtended by the greater chord, each being less than a semi-circumference. In the equal circles whose centers are C and M, arc AB > arc KL. |