# Equation of the form x+a=√(bx+c)

• MHB
• jennyyyyyy
In summary: Yes, we can answer part c of the question now.What would you do next?Well done!What would you do next?In summary, Jenny has been stuck on this problem for a few days, and she is hoping someone can help her. She has found that the equation $x+2=\sqrt{2x+67}$ can be solved for $x$ and that X=-9.
jennyyyyyy
i have been stuck
this problem for a few days now. could somebody help me please?

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?

anemone said:
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?

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...

anemone said:
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?

Well done!

What would you do next?

anemone said:
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

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

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

## 1. What is the purpose of the equation x+a=√(bx+c)?

The equation x+a=√(bx+c) is used to solve for the value of x in a quadratic function. It is commonly referred to as the "square root method" and is used to find the x-intercepts of a parabola.

## 2. How do I solve the equation x+a=√(bx+c)?

To solve this equation, you must first isolate the square root term by subtracting a from both sides. Then, square both sides of the equation to eliminate the square root. Finally, solve for x by using the quadratic formula or by factoring the resulting quadratic equation.

## 3. Can this equation have more than one solution?

Yes, this equation can have two solutions. Since a square root can have both a positive and negative value, the resulting quadratic equation can have two solutions for x.

## 4. Is there a specific method for solving this type of equation?

Yes, as mentioned before, this equation is commonly solved using the "square root method." However, it can also be solved by using the quadratic formula or by factoring the resulting quadratic equation.

## 5. Can this equation be used in real-world applications?

Yes, this equation can be used to solve various real-world problems involving parabolic shapes, such as finding the maximum height of a ball thrown into the air or the optimal dimensions of a container with a square base and a fixed volume.

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