# Quadratic with variable buried in sqrt?

• irishetalon00
In summary, the conversation discusses the process of solving a quartic equation with coefficients that do not easily simplify. The suggestion is made to use a root-seeking algorithm to solve it numerically, as closed-form formulas for quartic equations are rarely used in practice.
irishetalon00

## Homework Statement

0 = -250*(sqrt(x^2+1)-0.8)^2 + 98.1*x + 59.05

## The Attempt at a Solution

I can't figure out how to solve this equation! After expanding the squared term, I still end up with a sqrt(x^2+1) and I can't figure out how to perform a substitution or something to make it a nice clean quadratic.

Get the sqrt term alone on one side of the equation and then square both sides to get rid of the sqrt. You'll then have a quartic but if you simplify it as much as possible, a way to solve it may become apparent.

Ok I liked that suggestion. Now I'm here:
0 = -149809 + x(-19806.4) + x^2(-99901.4) + x^3(49050) + x^4(62500)
so now I have a quartic equation. Not sure how to proceed.
I mean it should reduce to a quadratic somehow, right?

It won't necessarily reduce to a quadratic. Sometimes one can solve a quartic by writing it as a quadratic in ##y\equiv x^2##, but this doesn't immediately appear to be one of those.

Are you required to solve it analytically? If not, the easiest thing is to solve it numerically by a root-seeking algorithm.
The coefficients don't look nice and neat like those of something that one would be given to solve analytically.

irishetalon00 said:
Ok I liked that suggestion. Now I'm here:
0 = -149809 + x(-19806.4) + x^2(-99901.4) + x^3(49050) + x^4(62500)
so now I have a quartic equation. Not sure how to proceed.
I mean it should reduce to a quadratic somehow, right?

No, not right. A quartic is a separate type of equation from a quadratic. Yours has two real and two complex roots. The two real roots are also the two real roots of your original equation, before you did any squaring.

Even though there are closed-form formulas for the solution of quartic equations, these formulas are rarely used in practice; numerical methods are much more frequently used.

## 1. What is a "Quadratic with variable buried in sqrt"?

A quadratic with variable buried in sqrt refers to a quadratic equation where the variable is inside a square root (sqrt) function. This type of equation is also known as a radical equation.

## 2. How do you solve a quadratic with variable buried in sqrt?

To solve a quadratic with variable buried in sqrt, you need to isolate the square root term on one side of the equation and then square both sides to eliminate the radical. This process may need to be repeated multiple times until the variable is no longer inside the square root function.

## 3. Can you use the quadratic formula to solve a quadratic with variable buried in sqrt?

Yes, the quadratic formula can be used to solve a quadratic with variable buried in sqrt. However, it may result in a complex solution since the formula involves calculating the square root of a negative number.

## 4. What is the difference between a quadratic with variable buried in sqrt and a regular quadratic equation?

The main difference is that a quadratic with variable buried in sqrt requires an extra step of isolating and eliminating the square root term in order to solve for the variable. This makes the process more complex and may result in complex solutions.

## 5. Can a quadratic with variable buried in sqrt have more than one solution?

Yes, a quadratic with variable buried in sqrt can have more than one solution. This is because when you square both sides of the equation to eliminate the square root, you may end up with both a positive and a negative solution.

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