Solve For $x$ In The Equation $x^2 - 10x + 25 = 35$.A. \$x = 5 \pm 2 \sqrt{5}$[/tex\]B. $x = 5 \pm \sqrt{35}$C. $x = 10 \pm 2 \sqrt{5}$D. \$x = 10 \pm \sqrt{35}$[/tex\]

by ADMIN 179 views

=====================================================

Introduction

The given equation is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 1$, $b = -10$, and $c = 25 - 35 = -10$. To solve for $x$, we can use the quadratic formula, which is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step 1: Write Down the Quadratic Formula

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this case, $a = 1$, $b = -10$, and $c = -10$.

Step 2: Plug in the Values into the Quadratic Formula

Substituting the values of $a$, $b$, and $c$ into the quadratic formula, we get $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(-10)}}{2(1)}$.

Step 3: Simplify the Expression

Simplifying the expression, we get $x = \frac{10 \pm \sqrt{100 + 40}}{2}$.

Step 4: Further Simplify the Expression

Further simplifying the expression, we get $x = \frac{10 \pm \sqrt{140}}{2}$.

Step 5: Simplify the Square Root

Simplifying the square root, we get $x = \frac{10 \pm 2\sqrt{35}}{2}$.

Step 6: Simplify the Expression

Simplifying the expression, we get $x = 5 \pm \sqrt{35}$.

Conclusion

Therefore, the solution to the equation $x^2 - 10x + 25 = 35$ is $x = 5 \pm \sqrt{35}$.

Answer

The correct answer is B. $x = 5 \pm \sqrt{35}$.

Discussion

The given equation is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 1$, $b = -10$, and $c = 25 - 35 = -10$. To solve for $x$, we can use the quadratic formula, which is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Substituting the values of $a$, $b$, and $c$ into the quadratic formula, we get $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(-10)}}{2(1)}$. Simplifying the expression, we get $x = \frac{10 \pm \sqrt{100 + 40}}{2}$. Further simplifying the expression, we get $x = \frac{10 \pm \sqrt{140}}{2}$. Simplifying the square root, we get $x = \frac{10 \pm 2\sqrt{35}}{2}$. Simplifying the expression, we get $x = 5 \pm \sqrt{35}$.

Related Topics

  • Quadratic Formula
  • Quadratic Equations
  • Algebra
  • Mathematics

References

=====================================================

Introduction

The given equation is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 1$, $b = -10$, and $c = 25 - 35 = -10$. To solve for $x$, we can use the quadratic formula, which is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this article, we will provide a Q&A section to help you understand the solution to the equation.

Q: What is the quadratic formula?

A: The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: How do I use the quadratic formula to solve for $x$?

A: To use the quadratic formula, you need to substitute the values of $a$, $b$, and $c$ into the formula. In this case, $a = 1$, $b = -10$, and $c = -10$.

Q: What is the solution to the equation $x^2 - 10x + 25 = 35$?

A: The solution to the equation is $x = 5 \pm \sqrt{35}$.

Q: Why do we need to simplify the square root?

A: We need to simplify the square root because it will make the solution easier to understand and work with.

Q: Can I use the quadratic formula to solve any quadratic equation?

A: Yes, you can use the quadratic formula to solve any quadratic equation in the form of $ax^2 + bx + c = 0$.

Q: What are the coefficients of the quadratic equation?

A: The coefficients of the quadratic equation are $a$, $b$, and $c$.

Q: How do I find the values of $a$, $b$, and $c$?

A: You can find the values of $a$, $b$, and $c$ by looking at the quadratic equation. In this case, $a = 1$, $b = -10$, and $c = -10$.

Q: Can I use the quadratic formula to solve for $x$ in a quadratic equation with complex solutions?

A: Yes, you can use the quadratic formula to solve for $x$ in a quadratic equation with complex solutions.

Q: What is the difference between a quadratic equation with real solutions and a quadratic equation with complex solutions?

A: A quadratic equation with real solutions has solutions that are real numbers, while a quadratic equation with complex solutions has solutions that are complex numbers.

Q: Can I use the quadratic formula to solve for $x$ in a quadratic equation with no real solutions?

A: Yes, you can use the quadratic formula to solve for $x$ in a quadratic equation with no real solutions.

Q: What is the difference between a quadratic equation with no real solutions and a quadratic equation with complex solutions?

A: A quadratic equation with no real solutions has no real solutions, while a quadratic equation with complex solutions has complex solutions.

Conclusion

In this article, we have provided a Q&A section to help you understand the solution to the equation $x^2 - 10x + 25 = 35$. We have also discussed the quadratic formula and how to use it to solve for $x$ in a quadratic equation.

Related Topics

  • Quadratic Formula
  • Quadratic Equations
  • Algebra
  • Mathematics

References