TANIM a, b, c ger\u00e7el say\u0131 ve a \u00b9 0 olmak \u00fczere,<\/p>\n
ax2 + bx + c = 0
\nbi\u00e7imindeki her a\u00e7\u0131k \u00f6nermeye ikinci dereceden bir bilinmeyenli denklem denir.
\n Bu a\u00e7\u0131k \u00f6nermeyi do\u011frulayan x say\u0131lar\u0131na denklemin k\u00f6kleri; t\u00fcm k\u00f6klerin olu\u015fturdu\u011fu k\u00fcmeye denklemin \u00e7\u00f6z\u00fcm k\u00fcmesi; \u00e7\u00f6z\u00fcm k\u00fcmesini bulmak i\u00e7in yap\u0131lan i\u015flemlere denklem \u00e7\u00f6zme; a, b, c say\u0131lar\u0131na da denklemin kat say\u0131lar\u0131 denir.<\/p>\n
B. \u0130K\u0130NC\u0130 DERECE DENKLEM\u0130N \u00c7\u00d6Z\u00dcM K\u00dcMES\u0130N\u0130N BULUNU\u015eU
\n1. \u00c7arpanlara Ay\u0131rma Y\u00f6ntemi
\n ax2 + bx + c = 0 denklemi f(x) . g(x) = 0
\n bi\u00e7iminde yaz\u0131labiliyorsa
\n f(x) = 0 veya g(x) = 0 olup \u00e7\u00f6z\u00fcm k\u00fcmesi;
\n \u00c7 = {x | x, f(x) = 0 veya Q(x) = 0 denklemini sa\u011flar} olur.
\n2. Diskiriminant (D) Y\u00f6ntemi
\n ax2 + bx + c = 0 denklemi a \u00b9 0 ve
\nD = b2 \u2013 4ac ise, \u00e7\u00f6z\u00fcm k\u00fcmesi
\nax2 + bx + c = 0
\n denkleminde, D = b2 \u2013 4ac olsun.
\na) D > 0 ise, denklemin farkl\u0131 iki ger\u00e7el k\u00f6k\u00fc vard\u0131r.
\nBu k\u00f6kleri,b) D < 0 ise, denklemin ger\u00e7el k\u00f6k\u00fc yoktur.\nc) D = 0 ise, denklemin e\u015fit iki ger\u00e7el k\u00f6k\u00fc vard\u0131r. \nBu k\u00f6kler,Denklemin bu k\u00f6klerine; e\u015fit iki k\u00f6k, \u00e7ak\u0131\u015f\u0131k k\u00f6k ya da \u00e7ift katl\u0131 k\u00f6k denir.\n\u00dc ax2 + bx + c = 0\n denkleminin k\u00f6kleri simetrik ise,\n1) b = 0 ve a \u00b9 0 d\u0131r.\n2) Simetrik k\u00f6kleri ger\u00e7el ise,\n b = 0, a \u00b9 0 ve a . c \u00a3 0 d\u0131r.\nC. \u0130K\u0130NC\u0130 DERECE DENKLEM\u0130N K\u00d6KLER\u0130 \u0130LE KATSAYILARI ARASINDAK\u0130 BA\u011eINTILAR\n ax2 + bx + c = 0 denkleminin k\u00f6kleri\n x1 ve x2 ise,\n \nD. K\u00d6KLER\u0130 VER\u0130LEN \u0130K\u0130NC\u0130 DERECEDEN DENKLEM\u0130N YAZILMASI\n K\u00f6kleri x1 ve x2 olan ikinci dereceden denklem;\n (x \u2013 x1) (x \u2013 x2) = 0 d\u0131r. Bu ifade d\u00fczenlenirse,\nx2 \u2013 (x1 + x2)x + x1x2 = 0 olur.\n\u00dc ax2 + bx + c = 0 ... (1) denkleminin k\u00f6kleri x1 ve x2 olsun. K\u00f6kleri mx1 + n ve\nmx2 + n olan ikinci dereceden denklem, (1) denkleminde x yerineyaz\u0131larak bulunur.\n\u00dc ax2 + bx + c = 0 ve dx2 + ex + f = 0 denklemlerinin \u00e7\u00f6z\u00fcm k\u00fcmeleri ayn\u0131 ise,\n\n\n\u00dc ax2 + bx + c = 0 ve dx2 + ex + f = 0\ndenklemlerinin sadece birer k\u00f6kleri e\u015fit ise,\n ax2 + bx + c = dx2 + ex + f\n (a \u2013 d)x2 + (b \u2013 e)x + c \u2013 f = 0 d\u0131r.\n Bu denklemin k\u00f6k\u00fc verilen iki denklemi de sa\u011flar.\n\u00dc\u00c7\u00dcNC\u00dc DERECEDEN DENKLEMLER\nA. TANIM\n a \u00b9 0 olmak \u00fczere, ax3 + bx2 + cx + d = 0 bi\u00e7imindeki denklemlere \u00fc\u00e7\u00fcnc\u00fc dereceden bir bilinmeyenli denklemler denir.\nB. \u00dc\u00c7\u00dcNC\u00dc DERECEDEN DENKLEM\u0130N K\u00d6KLER\u0130 \u0130LE KATSAYILARI ARASINDAK\u0130 BA\u011eINTILAR\n a \u00b9 0 ve ax3 + bx2 + cx + d = 0 denkleminin k\u00f6kleri x1, x2 ve x3 olsun. Buna g\u00f6re,\n \nC. K\u00d6KLER\u0130 VER\u0130LEN \u00dc\u00c7\u00dcNC\u00dc\nDERECE DENKLEM\u0130N YAZILMASI\n K\u00f6kleri x1, x2 ve x3 olan \u00fc\u00e7\u00fcnc\u00fc derece denklem\n (x \u2013 x1) (x \u2013 x2) (x \u2013 x3) = 0 d\u0131r.\n Bu denklem d\u00fczenlenirse,\nx3 \u2013 (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x \u2013 x1x2x3 = 0\n olur.\n\u00dc ax3 + bx2 + cx + d = 0 denkleminin k\u00f6kleri\n x1, x2, x3 olsun.\n1) Bu k\u00f6kler aritmetik dizi olu\u015fturuyorsa,\n\nx1 + x3 = 2x2 dir.\n2) Bu k\u00f6kler geometrik dizi olu\u015fturuyorsa,\n3) Bu k\u00f6kler hem aritmetik hem de geometrik dizi olu\u015fturuyorsa,\n\nx1 = x2 = x3 t\u00fcr.\nn, 1 den b\u00fcy\u00fck pozitif tam say\u0131 olmak \u00fczere,\n\nanxn + an \u2013 1xn \u2013 1 + ... + a1x + a0 = 0\n<\/p>\n","protected":false},"excerpt":{"rendered":"
TANIM a, b, c ger\u00e7el say\u0131 ve a \u00b9 0 olmak \u00fczere, ax2 + bx + c = 0 bi\u00e7imindeki her a\u00e7\u0131k \u00f6nermeye ikinci dereceden bir bilinmeyenli denklem denir. Bu a\u00e7\u0131k \u00f6nermeyi do\u011frulayan x say\u0131lar\u0131na denklemin k\u00f6kleri; t\u00fcm k\u00f6klerin olu\u015fturdu\u011fu k\u00fcmeye denklemin \u00e7\u00f6z\u00fcm k\u00fcmesi; \u00e7\u00f6z\u00fcm k\u00fcmesini bulmak i\u00e7in yap\u0131lan i\u015flemlere denklem \u00e7\u00f6zme; a, b, c …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1404,1403],"tags":[7282,7283,7284],"class_list":["post-3086","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-2-ve-3-dereceden-denklemler","tag-carpanlara-ayirma-yontemi","tag-diskiriminant"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3086","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=3086"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3086\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3086"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3086"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}