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
128, 132, 123, 139, 113, 150
324, 385, 460, 549, 651, 769
Find the wrong number in the given number series.
3, 8, 18, 40, 78, 158- Find the wrong number in the given number series.
6, 12, 36, 108, 540, 3240 - Find the wrong number in the given number series.
2, 6, 14, 30, 62, 126 20, 95, 220, 275, 620, 895
14, 16, 22, 46, 172, 886
119, 200, 137, 182, 156, 164
1320, 1352, 1390, 1436, 1488, 1548
- In the given number series, find the wrong number.
4, 9, 19, 39, 79, 159, 319