# asqrt(a)+bsqrt(b)=183 and asqrt(b)+bsqrt(a)=182 Find 9/5(a+b) ?

Apr 5, 2018

73

#### Explanation:

Let $\sqrt{a} = x \implies {x}^{2} = a$
and $\sqrt{b} = y \implies {y}^{2} = b$

So, the altered equations now look like,
${x}^{3} + {y}^{3} = 183$ and ${x}^{2} y + {y}^{2} x = 182$
color(white)(wwwwwwwwwd $\implies x y \left(x + y\right) = 182$

We know the basic identity,
color(magenta)((x+y)^3=x^ 3 +y^ 3 +3(x^ 2 y+y^ 2 x)

Plugging in the "altered" equations,
$\implies {\left(x + y\right)}^{3} = 183 + 3 \left(182\right)$
$\implies {\left(x + y\right)}^{3} = 729$
=> color(blue)(x+y = 9

Now, plug $x + y = 9$ into $x y \left(x + y\right) = 182$
=>color(red)( xy = 182/9

Also, ${\left(x - y\right)}^{2} = {\left(\textcolor{b l u e}{x + y}\right)}^{2} - 4 \textcolor{red}{x y}$

$\implies {\textcolor{b l u e}{9}}^{2} - 4 \left(\textcolor{red}{\frac{182}{9}}\right)$

$\implies 81 - \frac{4 \left(182\right)}{9} = \frac{1}{9}$

$\implies x - y = \pm \frac{1}{3}$

Since we've found out $x + y = 9$ and $x - y = \pm \frac{1}{3}$, we can find the values of $x$ and $y$ by elimination method.

When $x - y = + \frac{1}{3}$ color(white)(wwwwwwwwwd When $x - y = - \frac{1}{3}$

Adding both the equations, color(white)(wwwwwd Adding both the equations,

$2 x = 9 + \frac{1}{3}$ color(white)(wwwwwwwwwwwwwwd $2 x = 9 - \frac{1}{3}$

$x = \frac{14}{3}$ color(white)(wwwwwwwwwwwwwwwwww$x = \frac{13}{3}$

therefore, color(white)(wwwwwwwwwwwwwwwwd therefore,

$y = \frac{13}{3}$ color(white)(wwwwwwwwwwwwwwwwwd $y = \frac{14}{3}$

Either way, we have to find, $\frac{9}{5} \left(a + b\right) \implies \frac{9}{5} \left({x}^{2} + {y}^{2}\right)$

$\implies \frac{9}{5} \left({\left(\frac{13}{3}\right)}^{2} + {\left(\frac{14}{3}\right)}^{2}\right)$

$\implies \frac{\cancel{9}}{5} \times \frac{169 + 196}{\cancel{9}}$

$\implies 73$

Apr 5, 2018

#### Explanation:

Let $\sqrt{a} = x \implies {x}^{2} = a$
and $\sqrt{b} = y \implies {y}^{2} = b$

So, the altered equations now look like,
${x}^{3} + {y}^{3} = 183$ and ${x}^{2} y + {y}^{2} x = 182$
color(white)(wwwwwwwwwd $\implies x y \left(x + y\right) = 182$

We know the basic identity,
color(magenta)((x+y)^3=x^ 3 +y^ 3 +3(x^ 2 y+y^ 2 x)

Plugging in the "altered" equations,
$\implies {\left(x + y\right)}^{3} = 183 + 3 \left(182\right)$
$\implies {\left(x + y\right)}^{3} = 729$
=> color(blue)(x+y = 9

Now, plug $x + y = 9$ into $x y \left(x + y\right) = 182$
=>color(red)( xy = 182/9

We know, ${\left(x + y\right)}^{2} = {x}^{2} + {y}^{2} + 2 x y$
$\implies {x}^{2} + {y}^{2} = {\left(x + y\right)}^{2} - 2 x y$
color(white)(wwwwwwi $= {9}^{2} - 2 \left(\frac{182}{9}\right)$

color(white)(wwwwwwi $= \frac{365}{9}$

We need to find, $\frac{9}{5} \left(a + b\right) \implies \frac{9}{5} \left({x}^{2} + {y}^{2}\right)$

$\implies \frac{9}{5} \left(\frac{365}{9}\right)$

$\implies \frac{\cancel{9}}{5} \times \frac{365}{\cancel{9}}$

$\implies 73$

color(white)(wwwwwwi

color(white)(wwwwwwi

color(white)(wwwwwwi

Another way could be,
since,
x³+y³=183

(x+y)(x²+y²−xy)=183

(x+y)(x²+y²)−xy(x+y)=183

(x+y)(x²+y²)=182+183 =365

(x²+y²)=365/9
.
Then the required expression = (9/5)(a+b)=(9/5)(x²+y²)
$= \left(\frac{9}{5}\right) \left(3659\right) = 73$