Education

Solving quadratic Equations

A quadratic equation is an equation in which the highest power of the variable is 2. The general form of a quadratic equation is:
ax^2 + bx + c = 0
where a, b, and c are constants, and x is the variable.

  To solve a quadratic equation, there are several methods you can use:

• • Factoring: This involves finding two numbers that multiply to give the constant term c and add to give the coefficient of the x-term b. Once you have found these numbers, you can use them to write the quadratic equation in the form: (x + p)(x + q) = 0 where p and q are the two numbers you found. Then, you can solve for x by setting each factor equal to zero and solving for x.
  • Completing the square: This involves transforming the quadratic equation into a perfect square trinomial. To do this, add and subtract (b/2a)^2 to the left side of the equation, and then factor the resulting expression as a square: a(x + b/2a)^2 + c - b^2/4a = 0

Then, solve for x by taking the square root of both sides of the equation.

• Quadratic formula: This is a formula that can be used to solve any quadratic equation, regardless of whether it is factorable or not: x = (-b ± sqrt(b^2 - 4ac)) / 2a

To use this formula, simply substitute the values of a, b, and c into the formula and solve for x. These are the basic methods for solving quadratic equations. Practice and familiarity with these methods will help you become more confident and proficient at solving quadratic equations. Here are a few examples of quadratic equations and how to solve them:

• Solve the equation: x^2 - 4x - 5 = 0 To solve this equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a In this case, a = 1, b = -4, and c = -5. Substituting these values into the formula, we get: x = (4 ± sqrt((-4)^2 - 4(1)(-5))) / 2(1) x = (4 ± sqrt(36)) / 2 x = (4 ± 6) / 2 So the solutions are x = 5 and x = -1.
• Solve the equation: 3x^2 + 2x - 1 = 0 To solve this equation, we can use the quadratic formula again: x = (-b ± sqrt(b^2 - 4ac)) / 2a In this case, a = 3, b = 2, and c = -1. Substituting these values into the formula, we get: x = (-2 ± sqrt(2^2 - 4(3)(-1))) / 2(3) x = (-2 ± sqrt(28)) / 6 x = (-2 ± 2sqrt(7)) / 6 So the solutions are x = (-1 + sqrt(7)) / 3 and x = (-1 - sqrt(7)) / 3.
• Solve the equation: 2x^2 - 5x + 2 = 0 To solve this equation, we can try factoring: 2x^2 - 5x + 2 = (2x - 1)(x - 2) = 0 Setting each factor equal to zero, we get: 2x - 1 = 0 or x - 2 = 0 So the solutions are x = 1/2 and x = 2.

These are just a few examples of how to solve quadratic equations. Remember to always check your solutions by plugging them back into the original equation to make sure they work.