Solve for x in the equation x2-10+25=35

Mathematics

2 answers:

mina [271]2 years ago

7 0

X2-10+25=35
+10. +10
x2+25=45
-25 -25
x2=20
———
2
x=10
Tried my best

777dan777 [17]2 years ago

4 0

Hey there!

<h2> $\bold{x^2-10+25=35}$ </h2><h3>Firstly, we have to to simplify both sides of your equation </h3><h3></h3>

$\bold{x^2-10+25=35}$
$\bold{-10+25=15}$
Firstly, now that we got that portion out the way, let's $\bold{subtract}$ by the number $\bold{15}$ on each of your sides
$\bold{x^2+15-15=35-15}$
$\bold{Cancel:15-15}$
$\bold{Keep:35-15}$
$\bold{35-15=20}$
We get: $\bold{x^2=20}$
Take the square root of the number
$\bold{x=}$ ± $\bold{\sqrt{20}}$
$\boxed{\boxed{\bold{Answer:x=2\sqrt{5}[or]x=-2\sqrt{5}}}}$ $\checkmark$

<h3>Good luck on your assignment and enjoy your day!</h3><h3>~ $\frak{LoveYourselfFirst:)}$ </h3>

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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .