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If sin Q and cos Q are roots of equation (2p+1)x^-7p-3=0 has equal roots |
| Answer» The given quadratic equation is:\xa0(2p + 1)x2 - (7p + 2)x + 7p - 3 = 0Here, a = (2p + 1), b = -(7p + 2) and c = 7p - 3We know that, D = b2 - 4ac= [-(7p + 2)]2 - 4 {tex}\\times{/tex} (2p + 1) {tex}\\times{/tex} (7p - 3)= 49p2 + 4 + 28p - 4(14p2 - 6p + 7p - 3)= 49p2 + 4 + 28p - 4(14p2 + p - 3)= 49p2 + 4 + 28p - 56p2 - 4p + 12= -7p2 + 24p + 16Since it is given that the given equation has real and equal roots, so D = 0 i.e.,{tex}\\Rightarrow{/tex}\xa0-7p2 + 24p + 16 = 0{tex}\\Rightarrow{/tex}\xa07p2 - 24p - 16 = 0{tex}\\Rightarrow{/tex}\xa07p2 + 4p - 28p - 16 = 0{tex}\\Rightarrow{/tex}\xa0p(7p + 4) - 4(7p + 4) = 0{tex}\\Rightarrow{/tex}\xa0(p - 4)(7p + 4) = 0{tex}\\Rightarrow{/tex}\xa0p - 4 = 0 or 7p + 4 = 0{tex}\\Rightarrow{/tex}\xa0p = 4 or {tex}p = - \\frac{4}{7}{/tex}Therefore, after substituting the value of p in the given equation, the two equations will be:9x2 - 30x + 25 = 0 [For p = 4]{tex}\\Rightarrow{/tex}\xa09x2 - 15x - 15x + 25 = 0{tex}\\Rightarrow{/tex}\xa03x (3x - 5) - 5x (3x - 5) = 0{tex}\\Rightarrow{/tex}\xa0(3x - 5)2 = 0{tex}x = \\frac{5}{3}{/tex}or\xa0{tex}\\left( {2 \\times \\left( { - \\frac{4}{7}} \\right) + 1} \\right){x^2} - \\left( {7\\left( { - \\frac{4}{7}} \\right) + 2} \\right)x{/tex}{tex} + 7\\left( { - \\frac{4}{7}} \\right) - 3 = 0{/tex}\xa0[For {tex}p = - \\frac{4}{7}{/tex}]On solving we get,x2 - 14x + 49 = 0{tex}\\Rightarrow{/tex}\xa0x2 - 7x - 7x + 49 = 0{tex}\\Rightarrow{/tex}\xa0x (x - 7) - 7 (x - 7)= 0{tex}\\Rightarrow{/tex} (x - 7)2\xa0= 0{tex}\\Rightarrow{/tex} x = 7Thus,\xa0{tex}x = \\frac{5}{3}{/tex} or\xa0x = 7 | |