Let a, b, c and d be four pairwise distinct real numbers. If a and b are solutions of the equation x2−10cx−11d=0, and c and d are solutions of the equation x2−10ax−11b=0, determine the sum a+b+c+d.
Solution
From Vieta's formulae we have a+b=10c and c+d=10a. By adding these equations we get a+b+c+d=10(a+c). Since a is a solution of the equation x2−10cx−11d=0, and d=10a−c, it follows that 0=a2−10ac−11d=a2−10ac−11(10a−c)=a2−110a+11c−10ac. Analogously, we get c2−110c+11a−10ac=0. By subtracting these equations it follows that (a−c)(a+c−121)=0. Since a=c, we get a+c=121. Therefore, a+b+c+d=10⋅121=1210.
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Source: MathNet,
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