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Class-X (Mathematics)
        
 1.
          Squaring both sides
       x2 = 12 + x
      x2 - x - 12 = 0
     (x - 4)(x + 3) = 0
     x = 4, x = -3
 But  -3 is not possible as it is square root function, therefore x = 12 Ans
2. Solve using quadratic formula: 4x2 + 2(b - 3a)x - 3ab = 0
        Here D = [2(b - 3a)]2 - 4(4)(-3ab)
                      = 4b2 + 36a2 - 24ab + 48ab
                      = 4b2 + 36a2 + 24ab
                      = [2(b + 3a)]2
 
                - 2(b - 3a) ± sq.rt.[2(b + 3a)]2            - 2(b - 3a) ± 2(b + 3a)]
        x  = -----------------------------------------   =   --------------------------------
                                   2(4)                                                       8
         x = 3a/2 and  -b/2
 
3. PQR is a rt. triangle ang.Q  If QS=SR , show that PR2 = 4 PS2 - 3 PQ2 

In triangle PQR, PR2 = PQ2 + QR2
                                      = PQ2 + (2SQ)2 = PQ2 + 4SQ2
                                      = PQ2 + 4(PS2 - PQ2)          {Using pythagoras theorem in triangle PQS}
                                PR2 = 4 PS2 - 3 PQ2
                                                                            Hence Proved