Divide 29 into two parts such that the sum of | vedclass

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Question

(A) Let the two parts be $x$ and $(29 - x)$.According to the problem,the sum of their squares is $425$:$x^2 + (29 - x)^2 = 425$$x^2 + (841 - 58x + x^2

Vedclass · Accepted Answer

$13, 16$

Vedclass · Answer

(A) Let the two parts be $x$ and $(29 - x)$.According to the problem,the sum of their squares is $425$:$x^2 + (29 - x)^2 = 425$$x^2 + (841 - 58x + x^2) = 425$$2x^2 - 58x + 841 - 425 = 0$$2x^2 - 58x + 416 = 0$Dividing by $2$:$x^2 - 29x + 208 = 0$Factoring the quadratic equation:$x^2 - 13x - 16x + 208 = 0$$x(x - 13) - 16(x - 13) = 0$$(x - 13)(x - 16) = 0$So,$x = 13$ or $x = 16$.If $x = 13$,the other part is $29 - 13 = 16$.If $x = 16$,the other part is $29 - 16 = 13$.Thus,the two parts are $13$ and $16$.

Divide $29$ into two parts such that the sum of their squares is $425$.

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