MHB Equation of the form x+a=√(bx+c)

[jennyyyyyy]

i have been stuck

this problem for a few days now. could somebody help me please?

 

[anemone]

Hi Jenny! Welcome to MHB!

For part a, let's say for $x=7$ and we set $a=2$, we know $\sqrt{bx+c}$ must be in the form $\sqrt{81}$ so that $7+2=9=\sqrt{81}$.

Since $b>1$, we can let $b$ to be a small positive number, says, $b=2$. The value of $c$ can be found out by doing the following:

$\begin{align*}2(7)+c&=81\\14+c&=81\\ \therefore c &=81-14\\&=67\end{align*}$

Can you answer part b of the question if the equation we just set up is $x+2=\sqrt{2x+67}$?

Don't worry, we can guide you through all parts of the problem, just that we need to work together so you can understand the problem better, okay?

 

[jennyyyyyy]

Hi Jenny! Welcome to MHB!

For part a, let's say for $x=7$ and we set $a=2$, we know $\sqrt{bx+c}$ must be in the form $\sqrt{81}$ so that $7+2=9=\sqrt{81}$.

Since $b>1$, we can let $b$ to be a small positive number, says, $b=2$. The value of $c$ can be found out by doing the following:

$\begin{align*}2(7)+c&=81\\14+c\\&=81\\ \therefore c\\&=81-14\\&=67\end{align*}$

Can you answer part b of the question if the equation we just set up is $x+2=\sqrt{2x+67}$?

Don't worry, we can guide you through all parts of the problem, just that we need to work together so you can understand the problem better, okay?

thank you so much! we just need to do the same step as a right?

 

[anemone]

Not really, solving means you need to solve for the value(s) of $x$, based on the values of $a,\,b$ and $c$ we got...

Can you solve $x+2=\sqrt{2x+67}$? You need to square both sides of the equation for a start...

 

[jennyyyyyy]

Hi Jenny! Welcome to MHB!

For part a, let's say for $x=7$ and we set $a=2$, we know $\sqrt{bx+c}$ must be in the form $\sqrt{81}$ so that $7+2=9=\sqrt{81}$.

Since $b>1$, we can let $b$ to be a small positive number, says, $b=2$. The value of $c$ can be found out by doing the following:

$\begin{align*}2(7)+c&=81\\14+c\\&=81\\ \therefore c\\&=81-14\\&=67\end{align*}$

Can you answer part b of the question if the equation we just set up is $x+2=\sqrt{2x+67}$?

Don't worry, we can guide you through all parts of the problem, just that we need to work together so you can understand the problem better, okay?

wait wait, i get it. sorry, i got confused a bit xD
(x+2)^2 = (√ 2x+67)^2
x^2 + 4x +4 = 2x+67
is it correct so far?

 

[anemone]

Well done!

What would you do next?

 

[jennyyyyyy]

Well done!

What would you do next?

(x+2)^2 = (√ 2x+67)^2
x^2 + 4x +4 = 2x+67
X^2 + 4x + 4 – 67 = 2x+67 –67
X^2+ 4x – 63 = 2x
X^2 + 4x – 63 – 2x = 2x – 2x
X^2 +2x – 63 = 0
X^2 + 9x – 7x – 63 = 0
X(x+9)-7(x+9)=0
(x+9) = 0
X-7= 0
X=-9
X=7

i solved it. i don't know if this correct though

 

[anemone]

Very good job! So, that is the answer for part b of the problem.

Moving on to part c, can you answer it now?