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Quadratic Equations

Long Answer Type

261. If the roots of the equation: (c² – ab)x² – 2(a² – bc)x + b² – ac = 0 are equal, then prove that either a = 0 or a³ + c³ = 3abc.

195 Views

262. Find the value of ‘K’ for which the equation Kx² + Kx + 1 = –4x² – x has real and equal roots.

Solution not provided.
Ans. K = 5, -3

160 Views

Short Answer Type

263. The sum ‘S’ of first n natural numbers is given by the relation

space straight S equals fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction

Find n, if the sum is 276.

88 Views

264. If the roots of the equation (a – b)x² + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

76 Views

265. A dealer sells an article for Rs. 75 and gains as much percent as the cost price of the article. Find its cost price.

84 Views

Long Answer Type

266. Solve the following equation by quadratic formula : (i) 2x⁴ – 5x² + 3 = 0 (ii) 2x⁴ – 5x^-2= 3, x ≠ 0.

79 Views

267. Find the roots of the following equation:

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82 Views

268. Find the roots of the following equation:

open parentheses fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction close parentheses to the power of 4 minus 10 open parentheses fraction numerator 2 straight x plus 1 over denominator straight x minus 1 end fraction close parentheses squared plus 9 equals 0

78 Views

Short Answer Type

269. Find the value of ‘K’ so that the sum of the roots of the equation 3x² (2K + 1 )x – K – 5 = 0 is equal to the product of the roots.

108 Views

270. The perimeter of a rectangular plot is 42 metres and its area is 108 m² Find the dimensions of the plot.

108 Views

Previous Year Papers

Test Series

Test Yourself

Long Answer Type

262. Find the value of ‘K’ for which the equation Kx2 + Kx + 1 = –4x2 – x has real and equal roots.

Short Answer Type

Long Answer Type

Short Answer Type

262. Find the value of ‘K’ for which the equation Kx² + Kx + 1 = –4x² – x has real and equal roots.