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Let the smaller integer be x. Then x(x+1)=306, so x²+x−306=0=(x−17)(x+18). Thus x=17.
The numbers are roots of x²−27x+182=0=(x−13)(x−14).
Length=2x+1. Therefore x(2x+1)=528, giving 2x²+x−528=0.
x(x+1)=156 gives x²+x−156=0=(x−12)(x+13). Thus the integers are 12 and 13.
Let breadth=x. Then length=2x. Hence 2x²=800, so x=20 m.
l+b=40 and lb=400. Therefore l=b=20 m.
Present ages are x and x+26. Three years later they are x+3 and x+29. Thus (x+3)(x+29)=360, giving x²+32x−273=0.
480/(x−8)=480/x+3. Simplifying gives x²−8x−1280=0.
Let shorter side=x and longer side=x+7. Then x²+(x+7)²=169. The positive solution is x=5 cm.
x²=5x+24, so x²−5x−24=0.
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