![CLASS 10 MATHS Polynomials Topic : If sum of the Zeroes of the quadratic polynomial 0) (k+3)x2 + 6x+1=0 is (-2/3), the find value of k Solution : We know that, Sum of Zeroes of ax' + bx + C = 0 PRuudla, LSe b/a is )= (-2 -6 From given, Remember {k+3} For any quadratic (-6)(3) = (-2)(k+3) polynomial ೩x2 + bx +c=0 (-18)/(-2) =k+3 b Sum of Zeroes = a 9 = k+ 3 9-3 = K k = 6 MATHS TUITIONS PRAVINDRA WARASIGUDA, SECUNDERABAD] Concept Clarity Confidence | Practice Success Easg Solve Score High Leam Smart CLASS 10 MATHS Polynomials Topic : If sum of the Zeroes of the quadratic polynomial 0) (k+3)x2 + 6x+1=0 is (-2/3), the find value of k Solution : We know that, Sum of Zeroes of ax' + bx + C = 0 PRuudla, LSe b/a is )= (-2 -6 From given, Remember {k+3} For any quadratic (-6)(3) = (-2)(k+3) polynomial ೩x2 + bx +c=0 (-18)/(-2) =k+3 b Sum of Zeroes = a 9 = k+ 3 9-3 = K k = 6 MATHS TUITIONS PRAVINDRA WARASIGUDA, SECUNDERABAD] Concept Clarity Confidence | Practice Success Easg Solve Score High Leam Smart](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_339079_201d9448_1784919337770_sc.jpg?tenant=sc&referrer=pwa-sharechat-v2&f=770_sc.jpg)
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P RAVINDRA#➗10th గణితం #📖ఎడ్యుకేషన్✍ #🏆పోటీ పరీక్షల స్పెషల్ #👩💻టెట్/DSC ప్రత్యేకం #🎓జాబ్ ప్రిపరేషన్📚![ARITHMETIC PROCRESSION is called Arithmetic Progression (A.P), A series of the form here "a is first term and "d" is a, a + d, a + 2d, a + 3d, common difference Common difference, Znd term st term Or *d is found by Znd term or any two consecutive terms 3rd term th term of an AP 1)d (n an = a + Example The series is clearly in A.P -1, 3, 7, 11, with common difference d = 3 - (-1) = 4/ Sum of first n terms 2) [2a (2)[a + !] 1)d] (n - + Or of the AP is = where | = last term in the given A.P ದ If a, b, C are three consecutive numbers, Condition for three they are in AP if and only if then consecutive numbers | to be in A.P 2b = a + c Let total number of terms be N. term from the end nth Then nth term from the end [ _ (n - 1)d of an A.P a + (N - n)d KEY POINTS IMPORTANT FORMULAE an = ೩ + (n - 1)d A.P is increasing If d > 0 = decreasing [ = a + (n - 1)d Ifd < 0 AP is > Sn = (2) [2a (n 1)d] If d = 0 All terms are equal > (?) [a + [] Number of terms from a to Sn = (Ida ) +1 (1da ) +1 is n n = Understand the concept; Practice PRAVINDRA makes Practice regularly, Perfect! Score excellently! ARITHMETIC PROCRESSION is called Arithmetic Progression (A.P), A series of the form here "a is first term and "d" is a, a + d, a + 2d, a + 3d, common difference Common difference, Znd term st term Or *d is found by Znd term or any two consecutive terms 3rd term th term of an AP 1)d (n an = a + Example The series is clearly in A.P -1, 3, 7, 11, with common difference d = 3 - (-1) = 4/ Sum of first n terms 2) [2a (2)[a + !] 1)d] (n - + Or of the AP is = where | = last term in the given A.P ದ If a, b, C are three consecutive numbers, Condition for three they are in AP if and only if then consecutive numbers | to be in A.P 2b = a + c Let total number of terms be N. term from the end nth Then nth term from the end [ _ (n - 1)d of an A.P a + (N - n)d KEY POINTS IMPORTANT FORMULAE an = ೩ + (n - 1)d A.P is increasing If d > 0 = decreasing [ = a + (n - 1)d Ifd < 0 AP is > Sn = (2) [2a (n 1)d] If d = 0 All terms are equal > (?) [a + [] Number of terms from a to Sn = (Ida ) +1 (1da ) +1 is n n = Understand the concept; Practice PRAVINDRA makes Practice regularly, Perfect! Score excellently!](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_317670_9a4e052_1785149294267_sc.jpg?tenant=sc&referrer=feed-service&f=267_sc.jpg)
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