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If the sum of two numbers is 14 and multiplication is 40 , then find their difference.

Solution

Let the numbers are $x$ and $y$.

$\therefore \quad$ From the question,

$$\begin{array}{r} x+y=14 \\ x y=40 \end{array}$$

We know that

\begin{aligned} &(x-y)^{2}=(x+y)^{2}-4 \times y \\ \Rightarrow &(x-y)^{2}=(14)^{2}-4 \times 40 \\ \Rightarrow &(x-y)^{2}=196-160 \\ \Rightarrow &(x-y)^{2}=36 \\ \therefore & x-y=\sqrt{36}=6: \text { Ans. } \end{aligned}

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