Julia and Ed were asked to add twonumbers together. Julia, by mistake, subtracted the two numbers and gave the answer as 10. Instead of adding, Ed multiplied the two numbers and the result was 651. What was the correct total?
+26387 Julia and Ed were asked to add twonumbers together. Julia, by mistake, subtracted the two numbers and gave the answer as 10. Instead of adding, Ed multiplied the two numbers and the result was 651. What was the correct total?
Number one = a
Number two = b
\(\begin{array}{rcll} a-b &=& 10 \\ a\cdot b &=& 651 \end{array}\)
\(\begin{array}{lrcll} (1) & (a+b)^2 &=& a^2+2ab+b^2 \\ (2) & (a-b)^2 &=& a^2-2ab+b^2 \\ \hline \\ (1)-(2) & (a+b)^2 - (a-b)^2 &=& a^2+2ab+b^2 -(a^2-2ab+b^2)\\ & (a+b)^2 - (a-b)^2 &=& a^2+2ab+b^2 - a^2+2ab-b^2\\ & (a+b)^2 - (a-b)^2 &=& a^2-a^2+2ab+2ab +b^2 -b^2\\ & (a+b)^2 - (a-b)^2 &=& 2ab+2ab\\ & (a+b)^2 - (a-b)^2 &=& 4ab \qquad & | \qquad +(a-b)^2 \\ & (a+b)^2 &=& 4ab +(a-b)^2 \qquad & | \quad a\cdot b = 651 \quad a-b = 10 \\ & (a+b)^2 &=& 4\cdot 651 +(10)^2 \\ & (a+b)^2 &=& 2604 + 100 \\ & (a+b)^2 &=& 2704 \qquad | \qquad \sqrt{} \\ & a+b &=& \pm \sqrt{2704} \\ & a+b &=& \pm 52 \\\\ \hline \\ (1) & a+b &=& 52 \\ (2) & a-b &=& 10 \\ \hline \\ (1) + (2) & a+b + a-b &=& 52 + 10 \\ & a+ a + b -b &=& 52 + 10 \\ & a+ a &=& 52 + 10 \\ &2a &=& 62 \quad | \quad :2 \\ & \mathbf{a_1} &\mathbf{=} & \mathbf{31} \\ \hline \\ (1) & a+b &=& -52 \\ (2) & a-b &=& 10 \\ \hline \\ (1) + (2) & a+b + a-b &=& -52 + 10 \\ & a+ a + b -b &=& -52 + 10 \\ & a+ a &=& -42 \\ &2a &=& -42 \quad | \quad :2 \\ & \mathbf{a_2} &\mathbf{=}& \mathbf{-21} \\ \hline \\ & a-b &=& 10 \\ & a_1 - b &=& 10 \\ & 31 - b &=& 10 \\ & b &=& 31 -10 \\ & \mathbf{b_1}&\mathbf{=}& \mathbf{21} \\ \hline \\ & a-b &=& 10 \\ & a_2 - b &=& 10 \\ & -21 - b &=& 10 \\ & b &=& -21 -10 \\ & \mathbf{b_2} &\mathbf{=}& \mathbf{-31} \\ \end{array}\)
First correct total is \( (a+b) = 52\), because \(a_1 \cdot b_1 = 31\cdot 21 = 651\) and \(a_1 - b_1 = 31 - 21 = 10 \)
Second correct total is \((a+b) = -52\), because \(a_2 \cdot b_2 = -21\cdot( -31) = 651\) and \(a_2 - b_2 = -21 - (-31)= 31-21 = 10 \)
+26387 Julia and Ed were asked to add twonumbers together. Julia, by mistake, subtracted the two numbers and gave the answer as 10. Instead of adding, Ed multiplied the two numbers and the result was 651. What was the correct total?
Number one = a
Number two = b
\(\begin{array}{rcll} a-b &=& 10 \\ a\cdot b &=& 651 \end{array}\)
\(\begin{array}{lrcll} (1) & (a+b)^2 &=& a^2+2ab+b^2 \\ (2) & (a-b)^2 &=& a^2-2ab+b^2 \\ \hline \\ (1)-(2) & (a+b)^2 - (a-b)^2 &=& a^2+2ab+b^2 -(a^2-2ab+b^2)\\ & (a+b)^2 - (a-b)^2 &=& a^2+2ab+b^2 - a^2+2ab-b^2\\ & (a+b)^2 - (a-b)^2 &=& a^2-a^2+2ab+2ab +b^2 -b^2\\ & (a+b)^2 - (a-b)^2 &=& 2ab+2ab\\ & (a+b)^2 - (a-b)^2 &=& 4ab \qquad & | \qquad +(a-b)^2 \\ & (a+b)^2 &=& 4ab +(a-b)^2 \qquad & | \quad a\cdot b = 651 \quad a-b = 10 \\ & (a+b)^2 &=& 4\cdot 651 +(10)^2 \\ & (a+b)^2 &=& 2604 + 100 \\ & (a+b)^2 &=& 2704 \qquad | \qquad \sqrt{} \\ & a+b &=& \pm \sqrt{2704} \\ & a+b &=& \pm 52 \\\\ \hline \\ (1) & a+b &=& 52 \\ (2) & a-b &=& 10 \\ \hline \\ (1) + (2) & a+b + a-b &=& 52 + 10 \\ & a+ a + b -b &=& 52 + 10 \\ & a+ a &=& 52 + 10 \\ &2a &=& 62 \quad | \quad :2 \\ & \mathbf{a_1} &\mathbf{=} & \mathbf{31} \\ \hline \\ (1) & a+b &=& -52 \\ (2) & a-b &=& 10 \\ \hline \\ (1) + (2) & a+b + a-b &=& -52 + 10 \\ & a+ a + b -b &=& -52 + 10 \\ & a+ a &=& -42 \\ &2a &=& -42 \quad | \quad :2 \\ & \mathbf{a_2} &\mathbf{=}& \mathbf{-21} \\ \hline \\ & a-b &=& 10 \\ & a_1 - b &=& 10 \\ & 31 - b &=& 10 \\ & b &=& 31 -10 \\ & \mathbf{b_1}&\mathbf{=}& \mathbf{21} \\ \hline \\ & a-b &=& 10 \\ & a_2 - b &=& 10 \\ & -21 - b &=& 10 \\ & b &=& -21 -10 \\ & \mathbf{b_2} &\mathbf{=}& \mathbf{-31} \\ \end{array}\)
First correct total is \( (a+b) = 52\), because \(a_1 \cdot b_1 = 31\cdot 21 = 651\) and \(a_1 - b_1 = 31 - 21 = 10 \)
Second correct total is \((a+b) = -52\), because \(a_2 \cdot b_2 = -21\cdot( -31) = 651\) and \(a_2 - b_2 = -21 - (-31)= 31-21 = 10 \)
