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Class 10 Maths Chapter 2 (Hindi)

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Summary

This document explains polynomials, including linear, quadratic, and cubic polynomials, and their degrees. It covers the geometrical meaning of the roots of a polynomial equation. It is for a class 10 mathematics textbook.

Full Transcript

cgqin 13 cgqin 2 2.1 Hkwfedk d{kk IX esa] vkius ,d pj okys cgqinksa (polynomials) ,oa mudh ?kkrksa (degree) osQ ckjs esa vè;;u fd;k gSA ;k...

cgqin 13 cgqin 2 2.1 Hkwfedk d{kk IX esa] vkius ,d pj okys cgqinksa (polynomials) ,oa mudh ?kkrksa (degree) osQ ckjs esa vè;;u fd;k gSA ;kn dhft, fd pj x osQ cgqin p(x) esa x dh mPpre ?kkr (power) cgqin dh ?kkr (degree) dgykrh gSA mnkgj.k osQ fy,] 4x + 2 pj x esa ?kkr 1 dk cgqin gS] 2y2 – 3y + 4 pj y esa ?kkr 2 dk cgqin gS] 5x3 – 4x2 + x – 2 pj x esa ?kkr 3 dk 3 cgqin gS vkSj 7u6 – u 4  4u 2  u  8 pj u esa ?kkr 6 dk cgqin gSA O;atd 1 , x  2 , 1 2 x 1 bR;kfn cgqin ugha gSAa x + 2x + 3 2 ?kkr 1 osQ cgqin dks jSf[kd cgqin (linear polynomial) dgrs gSaA mnkgj.k osQ 2 2 fy,] 2x – 3, 3 x  5, y  2, x , 3z + 4, u  1 , bR;kfn lHkh jSf[kd cgqin gSaA 11 3 tcfd 2x + 5 – x2, x3 + 1, vkfn izdkj osQ cgqin jSf[kd cgqin ugha gSaA ?kkr 2 osQ cgqin dks f}?kkr cgqin (quadratic polynomial) dgrs gSaA f}?kkr s (quadrate) 'kCn ls cuk gS] ftldk vFkZ gS ^oxZ*A 2 x 2  3x  2 , (quadratic) 'kCn DokMªV 5 u 2 1 y2 – 2, 2  x 2  3 x,  2u 2  5, 5v 2  v, 4 z 2  , f}?kkr cgqinksa osQ oqQN mnkgj.k 3 3 7 gSa (ftuosQ xq.kkad okLrfod la[;k,¡ gSa)A vf/d O;kid :i esa] x esa dksbZ f}?kkr cgqin ax2 + bx + c, tgk¡ a, b, c okLrfod la[;k,¡ gSa vkSj a ¹ 0 gS] osQ izdkj dk gksrk gSA ?kkr 3 dk cgqin f=k?kkr cgqin (cubic polynomial) dgykrk gSA f=k?kkr cgqin osQ oqQN mnkgj.k gSa% 2 – x3, x3, 2 x3 , 3 – x2 + x3, 3x3 – 2x2 + x – 1 Rationalised 2023-24 14 xf.kr okLro esa] f=k?kkr cgqin dk lcls O;kid :i gS% ax3 + bx2 + cx + d, tgk¡ a, b, c, d okLrfod la[;k,¡ gSa vkSj a ¹ 0 gSA vc cgqin p(x) = x2 – 3x – 4 ij fopkj dhft,A bl cgqin esa x = 2 j[kus ij ge p(2) = 22 – 3 × 2 – 4 = – 6 ikrs gSaA x2 – 3x – 4 esa] x dks 2 ls izfrLFkkfir djus ls izkIr eku ^&6*] x2 – 3x – 4 dk x = 2 ij eku dgykrk gSA blh izdkj p(0), p(x) dk x = 0 ij eku gS] tks – 4 gSA ;fn p(x), x esa dksbZ cgqin gS vkSj k dksbZ okLrfod la[;k gS] rks p(x) esa x dks k ls izfrLFkkfir djus ij izkIr okLrfod la[;k p(x) dk x = k ij eku dgykrh gS vkSj bls p(k) ls fu:fir djrs gSaA p(x) = x2 –3x – 4 dk x = –1 ij D;k eku gS\ ge ikrs gSa % p(–1) = (–1)2 –{3 × (–1)} – 4 = 0 lkFk gh] è;ku nhft, fd p(4) =42 – (3 ´ 4) – 4 = 0 gSA D;ksfa d p(–1) = 0 vkSj p(4) = 0 gS] blfy, –1 vkSj 4 f}?kkr cgqin x2 – 3x – 4 osQ 'kwU;d (zeroes) dgykrs gSaA vf/d O;kid :i esa] ,d okLrfod la[;k k cgqin p(x) dk 'kwU;d dgykrh gS] ;fn p(k) = 0 gSA vki d{kk IX esa i

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