MAT135Y5 Lecture Notes - Lecture 1: Quadratic Formula

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The equation (cid:1853)2+(cid:1854)+(cid:1855)=0 (for (cid:1853) 0) has: no real solution if (cid:1854)2 4(cid:1853)(cid:1855)<0, unique solution if (cid:1854)2 4(cid:1853)(cid:1855)=0, given by x = (cid:3029)2(cid:3028, two real distinct solutions, if (cid:1854)2 4(cid:1853)(cid:1855)>0, given by = (cid:3029) (cid:3029)2 4(cid:3028)(cid:3030) , but (cid:4672)+ (cid:3029)2(cid:3028)(cid:4673)2 0 no solution: if (cid:1854)2 4(cid:1853)(cid:1855)<0 , then (cid:3029)2 4(cid:3028)(cid:3030) 4(cid:3028)2: if (cid:1854)2 4(cid:1853)(cid:1855)=0 , then (cid:4672)+ (cid:3029)2(cid:3028)(cid:4673)2=0 but this says + (cid:3029)2(cid:3028)=0 that is = (cid:3029)2(cid:3028, if (cid:1854)2 4(cid:1853)(cid:1855)>0, then can take square root of both sides (cid:4672)+ (cid:3029)2(cid:3028)(cid:4673)= (cid:3029)2 4(cid:3028)(cid:3030)

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MAT135Y5 Lecture Notes - Lecture 1: Quadratic Formula

MAT135Y5 Full Course Notes

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