Question
I). p2 - 20p + 51 = 0, II).
2q2 - 7q + 6 = 0 The following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give the answer.Solution
ATQ, p2 - 20p + 51 = 0 or p2 - 17p - 3p – 51 = 0 or, (p - 17) × (p - 3) = 0 or, p = 17, 3 2q2 - 7q + 6 = 0 Or 2q2 - 4q - 3q + 6 = 0 Or (q - 2) × (2q - 3) = 0 Or, q = 2, 3/2 So, p > q
If p = 24 - q - r and pq + r(q + p) = 132, then find the value of (p² + q² + r²).
((99.9 - 20.9)² + (99.9 + 20.9)² )/(99.9 x 99.9 + 20.9 x 20.9) = ?
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Find the value of the given expression-
(4x+4 -5× 4x+2) / 15×4x – 22×4x
If 4x² + y² = 40 and x y = 6, then find the value
of 2x + y?
If p = 40 - q - r and pq + r(q + p) = 432, then find the value of (p² + q² + r²).
47.98 × 4.16 + √325 × 12.91 + ? = 79.93 × 5.91
If x + y = 4 and (1/x) + (1/y) = 24/7, then the value of (x3 + y3).
- If p = 20 - q - r and pq + r(p + q) = 154, then find the value of (p² + q² + r²).
If a = (√2 - 1)1/3, then the value of (a-1/a)3 +3(a-1/a) is: