Question
If ‘y1’ and ‘y2’ are the roots of quadratic
equation 5y2 – 25y + 15 = 0, then find the quadratic equation whose roots are ‘3y1’ and ‘3y2’.Solution
For an equation of the form ax2 + bx + c = 0, Sum of roots = (-b/a) Product of roots = (c/a) Sum of roots (y1 + y2) = – (–25/5) = –(–5) = 5 Product of roots (y1 × y2) = (15/5) = 3 Sum of roots of required equation = 3y1 + 3y2 = 3 × (y1 + y2) = 3 × 5 = 15 Product of roots of required equation = 3y1 × 3y2 = 9 × y1 × y2 = 9 × 3 = 27 Quadratic equation is: y2 – (sum of roots) × y + (product of roots). So, required equation = y2 – 15y + 27 = 0
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