How do you insert parentheses so the statement 4+10+8-9•8= -50 is true?

Parenthesis aren't needed if PEMDAS is followed. However, parenthesis can be added to reinforce the need to multiply first.

Explanation:

Looking at the original, the key part is to get the 8 to multiply against the correct number so that it all works out to be $- 50$. And it turns out that we need it to multiply against $- 9$, so that:

$22 - 72 = - 50$

The way the equation is written right now, we'd have 4 adding to 10 adding to 8 and then subtracting out 72... which is exactly what we want. So my first thought is that parenthesis aren't needed. Without parenthesis, we get:

$4 + 10 + 8 - 9 \cdot 8 = - 50$

$4 + 10 + 8 - 72 = - 50$

$- 50 = - 50$

If we want to throw some parenthesis in here, then we'd add them so that it reads either as $\left(9 \cdot 8\right)$ - we need to keep the minus sign outside of the parenthesis because if it sits within them, it'll look like the $+ 8$ needs to multiply against the result of $- 9 \cdot 8$. So that would look like:

$4 + 10 + 8 - \left(9 \cdot 8\right) = - 50$

Jul 7, 2016

No need for parentheses

Explanation:

This problem can be formulated as a minimum determination.

Find

Minimum of

$f \left({x}_{1} , {x}_{2} , {x}_{3}\right) = {\left(4 {x}_{1} + 10 {x}_{2} + 8 {x}_{3} - 72 + 50\right)}^{2}$

subjected to

${x}_{1}^{2} = 1 , {x}_{2}^{2} = 1 , {x}_{3}^{2} = 1$

The lagrangian is

$L \left({x}_{1} , {x}_{2} , {x}_{3} , {\lambda}_{1} , {\lambda}_{2} , {\lambda}_{3}\right) = f \left({x}_{1} , {x}_{2} , {x}_{3}\right) + {\sum}_{k = 1}^{3} {\lambda}_{k} \left({x}_{k}^{2} - 1\right)$

The stationary points are the solutions to

$\nabla L \left({x}_{1} , {x}_{2} , {x}_{3} , {\lambda}_{1} , {\lambda}_{2} , {\lambda}_{3}\right) = \vec{0}$

or

{ (2 lambda_1 x_1 + 8 (-22 + 4 x1 + 10 x_2 + 8 x_3)=0), (2 lambda_2 x_2 + 20 (-22 + 4 x_1 + 10 x_2 + 8 x_3)=0), (2 lambda_3 x_3 + 16 (-22 + 4 x_1 + 10 x_2 + 8 x_3)=0), ( x_1^2-1), ( x_2^2-1), (x_3^2-1) :}

The solutions

( (x_1 = -1, x_2 = -1, x_3 = -1->1936), (x_1 = 1, x_2 = -1, x_3 = -1->1296), (x_1 = -1, x_2 = 1, x_3 = -1->576), (x_1 = 1, x_2 = 1, x_3 = -1->256), (x_1 = -1, x_2 = -1, x_3 = 1->784), (x_1 = 1, x_2 = -1, x_3 = 1->400), (x_1 = -1, x_2 = 1, x_3 = 1->64), (x_1 = 1, x_2 = 1, x_3 = 1->0))

The last column shows the $f \left({x}_{1} , {x}_{2} , {x}_{3}\right)$ corresponding values

The solution is

$\left({x}_{1} = 1 , {x}_{2} = 1 , {x}_{3} = 1\right)$ so no need for parentheses