Top 13+ (xiii) (k+1)x^(2)-2(3k+1)x+8k+1=0 hay nhất
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1.(K + 1)X2 - 2(3k + 1)X + 8k + 1 = 0 - Mathematics
The given quadric equation is (k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0, and roots are real and equal. Then find the value of k. Here,. a = k + 1, b = -2(3k + 1)x ...
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2.2(3k + 1)x + 8k + 1 = 0 Has Equal Roots - Quadratic Equations - Toppr
Click here👆to get an answer to your question ✍️ For what value/values of k : (k + 1)x^2 - 2(3k + 1)x + 8k + 1 = 0 has equal roots.
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3.Find The Values Of K For Roots Are Real And Equal In Equation: (k + 1)x^2
Find the values of k for roots are real and equal in equation: (k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0.
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4.(k+1)x^2-2(3k+1)x+8k+1=0 Find The Value Of K If The Eqation Has ...
here the equation has real and equal roots therefore, D = 0 this implies (-2(3k+1)) - 4(k+1)(8k+1) = 0 from this we get, 4k - 12k = 0
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5.Find The Values Of K For Which (k+1)x^2-2(3k+1)x+8k+1=0 Has Real ...
Find the values of k for which (k+1)x^2-2(3k+1)x+8k+1=0 has real and equal roots. Bị thiếu: (xiii) | Phải bao gồm: (xiii)
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6.Question And Answer - SnapSolve
(xiii) (k+1)x^{2}-2(3k+1)x+8k+1=0 ... are increased by 3 they are in the ratio2:3 2:3 Q29 Draw the the graph of the equationsx- y+1=0 and 3x+2y-12=0 ...
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7.(K+1)x^2-2(3k+1)x+8k+1=0 - Solution
Simple and best practice solution for (K+1)x^2-2(3k+1)x+8k+1=0 equation. Check how easy it is, and learn it for the future. Our solution is simple, ... Bị thiếu: (xiii) | Phải bao gồm: (xiii)
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8.For What Value Of K,4 K X2+2 K+4 X+8 K+1=0, Is A Perfect Square - Byju's
The given quadratic will be a perfect square if it has two real and equal roots. To have two real and equal roots the discriminant must be zero.
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9.Find The Value Of K For Which The Roots Are Real And Equal. (k+1)x^2-2 ...
Thời lượng: 5:09 Đã đăng: 28 thg 12, 2020 VIDEO
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10.Chapter 8: Quadratic Equations Exercise – 8.6 - AskIITians
(iii) For what value of k (4 - k)x2 + (2k + 4)x + (8k + 1) = 0 is a perfect square. (iv) Find the least positive value of k for which the equation x2 + kx + ...
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11.Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations
16 thg 5, 2021 · (xiii) (k + 1) x² – 2 (3k + 1) x + 8k + 1 = 0. Solution: Here a=k+1, b=-2(3k+1), c=8k+1. Discriminant(D)=b2-4ac. =[-2(3k+1)2-4*(k+1)(8k+1)].
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12.[PDF] RD SHARMA - Solutions
(iii) For what value of k (4-k)x² + (2k+4)x +(8k+1) =0 is a perfect square. ... The given equation (3k+1)x²+2(k+1)x+k=Ois in the form of ax²+bx+c=0 where a= ...
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13.RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Ex 8.6
(x) kx² + kx + 1 = – 4x² – x (xi) (k + 1) x² + 2 (k + 3) x + (k + 8) = 0 (xii) x² – 2kx + 7k – 12 = 0 (xiii) (k + 1) x² – 2 (3k + 1) x + 8k + 1 = 0
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