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7. The roots of each of the following quadratic equations are real and equal, find k.
(1) 3\(y^2\) + ky + l2 = 0

(2) kx (x – 2) + 6 = 0

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(1)    3\(y^2\) + ky + l2 = 0
Sol.  3\(y^2\) + ky + l2 = 0
        Comparing with a\(x^2 \) + bx + c = 0
        we get, a = 3, b = k and c = 12

       Let’s find value of discriminant
       Δ   =   D  =   \(b^2\) – 4ac
                      =   \(k^2\) – 4 × 3 × 12
                      =   \(k^2\) – 144                               ...(I)
       but roots of equation are real and equal
∴     Δ  =   D    =   0                        ...(II)
∴     \(k^2\) – 144  =   0               [From (I) & (II)]
∴     \(k^2\) – 122  =   0
      By using \(a^2\)\(b^2\) = (a + b) (a – b)
∴   (k + 12)  (k – 12)  =  0
∴   k + 12  =  0    or   k – 12  =  0
∴          k   = –12 or          k   = 12

 

(2)   kx (x – 2) + 6 = 0
Sol. kx (x – 2) + 6 = 0
∴     k\(x^2 \) – 2kx + 6 = 0
       comparing with a\(x^2 \) + bx + c = 0
∴    a = k, b = –2k and c = 6
      Let’s find value of discriminant
      Δ  =  D  =  \(b^2\) – 4ac
                   =  \((–2k)^2\) – 4 × k × 6
                   =  4\(k^2\) – 24k                             ...(I)
       But roots of equation are real and equal
∴    Δ   =   D     =  0                                 ...(II)
∴    4\(k^2\) – 24k  =  0                    [From (I) & (II)]
∴    4k(k – 6)    =  0
             k – 6    =  \(0\over 4k\)
∴           k – 6    =  0
∴                 k    =  6

uÉaÉïxÉqÉÏMüUhÉ August 10 , 2018 0 Comments 7 views