If $\sqrt{u} + \sqrt{v} - \sqrt{w} = 0$ $sqrt(u) + sqrt(v) - sqrt(w) = 0$ , then find the value of $(u + v - w)$ $(u+v-w)$ ?

Question

If $\sqrt{u} + \sqrt{v} - \sqrt{w} = 0$ $sqrt(u) + sqrt(v) - sqrt(w) = 0$ , then find the value of $(u + v - w)$ $(u+v-w)$ .

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Answer 1 · 2018-07-17T17:50:45

See a solution process below:

First, solve for $w$ $w$

$\sqrt{u} + \sqrt{v} - \sqrt{w} = 0$ $sqrt(u) + sqrt(v) - sqrt(w) = 0$

$\sqrt{u} + \sqrt{v} - \sqrt{w} + \sqrt{w} = 0 + \sqrt{w}$ $sqrt(u) + sqrt(v) - sqrt(w) + sqrt(w) = 0 + sqrt(w)$

$\sqrt{u} + \sqrt{v} - 0 = \sqrt{w}$ $sqrt(u) + sqrt(v) - 0 = sqrt(w)$

$\sqrt{u} + \sqrt{v} = \sqrt{w}$ $sqrt(u) + sqrt(v) = sqrt(w)$

${(\sqrt{u} + \sqrt{v})}^{2} = {(\sqrt{w})}^{2}$ $(sqrt(u) + sqrt(v))^2 = (sqrt(w))^2$

${(\sqrt{u})}^{2} + 2 \sqrt{u} \sqrt{v} + {(\sqrt{v})}^{2} = w$ $(sqrt(u))^2 + 2sqrt(u)sqrt(v) + (sqrt(v))^2 = w$

$u + 2 \sqrt{u} \sqrt{v} + v = w$ $u + 2sqrt(u)sqrt(v) + v = w$

Substituting the left side of the equation for the $w$ $w$ in the expression gives:

$(u + v - w)$ $(u + v - w)$ becomes:

$(u + v - (u + 2 \sqrt{u} \sqrt{v} + v)) \Rightarrow$ $(u + v - (u + 2sqrt(u)sqrt(v) + v)) =>$

$(u + v - u - 2 \sqrt{u} \sqrt{v} - v) \Rightarrow$ $(u + v - u - 2sqrt(u)sqrt(v) - v) =>$

$(u - u + v - v - 2 \sqrt{u} \sqrt{v}) \Rightarrow$ $(u - u + v - v - 2sqrt(u)sqrt(v)) =>$

$(0 + 0 - 2 \sqrt{u} \sqrt{v}) \Rightarrow$ $(0 + 0 - 2sqrt(u)sqrt(v)) =>$

$- 2 \sqrt{u} \sqrt{v}$ $-2sqrt(u)sqrt(v)$