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| 1. |
Find zeroes of p(x) = abx2 + ( b2 - ac)x - bc |
| Answer» We have,f(x) = abx2 + (b2 - ac)x - bc= abx2 + b2x - acx - bc= bx(ax + b ) - c(ax + b)= (ax + b) (bx - c)Now r(x)=0 if\xa0ax+b=0 or bx-c=0{tex}\\style{font-family:Arial}{\\begin{array}{l}\\style{font-size:12px}{\\mathrm i}\\style{font-size:12px}.\\style{font-size:12px}{\\mathrm e}\\style{font-size:12px}.\\style{font-size:12px}\\;\\style{font-size:12px}{\\mathrm x}\\style{font-size:12px}=\\style{font-size:12px}-\\frac{\\style{font-size:12px}{\\mathrm b}}{\\style{font-size:12px}{\\mathrm a}}\\style{font-size:12px}\\;\\style{font-size:12px}{\\mathrm{or}}\\style{font-size:12px}\\;\\style{font-size:12px}{\\mathrm x}\\style{font-size:12px}=\\frac{\\style{font-size:12px}{\\mathrm c}}{\\style{font-size:12px}{\\mathrm b}}\\\\\\end{array}}{/tex}Thus, the zeroes of f(x) are :\xa0{tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\mathrm\\alpha}\\style{font-size:12px}=\\style{font-size:12px}-\\frac{\\style{font-size:12px}{\\mathrm b}}{\\style{font-size:12px}{\\mathrm a}}\\style{font-size:12px}\\;\\style{font-size:12px}{\\mathrm{and}}\\style{font-size:12px}\\;\\style{font-size:12px}{\\mathrm\\beta}\\style{font-size:12px}=\\frac{\\style{font-size:12px}{\\mathrm c}}{\\style{font-size:12px}{\\mathrm b}}}{/tex} | |