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Find value of p for which the quadratic equation (2p+1)xsq.-(7p+2)x+7P-3=0 |
| Answer» The given quadratic equation is:\xa0(2p + 1)x2 - (7p + 2)x + 7p - 3 = 0Here, a = (2p + 1), b = -1(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} -7p2 + 24p + 16 = 0{tex}\\Rightarrow{/tex} 7p2 - 24p - 16 = 0{tex}\\Rightarrow{/tex} 7p2 + 4p - 28p - 16 = 0{tex}\\Rightarrow{/tex} p(79 + 4) - 4(7p + 4) = 0{tex}\\Rightarrow{/tex} (p - 4)(7p + 4) = 0{tex}\\Rightarrow{/tex} p - 4 = 0 or 7p + 4 = 0{tex}\\Rightarrow{/tex} p = 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 = 0or\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}After solving the second equation, the two equations can be rewritten as, 9x2 - 30x + 25 = 0 or x2 - 14x + 49 = 0After solving these two equations by factorization method, we have\xa0{tex}x = \\frac{5}{3}{/tex} or x = 7. | |