Welcome to our lesson on solving quadratic equations. A quadratic equation has the form a x squared plus b x plus c equals zero, where a, b, and c are constants, and a is not equal to zero. There are three main methods to solve quadratic equations: factoring, using the quadratic formula, and completing the square. On the right, you can see a graph of a quadratic function y equals x squared minus 2. The x-intercepts, or roots, of this function are the solutions to the equation x squared minus 2 equals zero, which are x equals negative square root of 2 and x equals positive square root of 2.
Let's look at our first method: factoring. Factoring works when the quadratic expression can be written as a product of two linear factors. The general form a x squared plus b x plus c can be factored as a times x minus r1 times x minus r2, where r1 and r2 are the roots of the equation. To solve by factoring, first ensure the equation is in standard form with zero on one side. Then try to factor the expression into a product of linear terms. Next, set each factor equal to zero, and finally solve for x. Let's see an example: x squared plus 5x plus 6 equals zero. We can factor this as x plus 2 times x plus 3 equals zero. Setting each factor equal to zero, we get x plus 2 equals zero or x plus 3 equals zero. Solving for x gives us x equals negative 2 or x equals negative 3. These are our solutions.
Our second method is using the quadratic formula. This formula works for any quadratic equation and gives the solutions directly. For an equation in the form a x squared plus b x plus c equals zero, the solutions are given by x equals negative b plus or minus the square root of b squared minus 4ac, all divided by 2a. The expression under the square root, b squared minus 4ac, is called the discriminant. It tells us about the nature of the solutions. If the discriminant is positive, we have two distinct real solutions. If it's zero, we have one repeated real solution. And if it's negative, we have two complex solutions. Let's solve an example: 2x squared minus 4x minus 3 equals zero. First, we identify the coefficients: a equals 2, b equals negative 4, and c equals negative 3. Next, we substitute these values into the quadratic formula. After simplifying, we get x equals 1 plus or minus the square root of 10 divided by 2. This gives us two solutions: approximately 2.58 and negative 0.58.
Our third method is completing the square. This technique transforms the quadratic expression into a perfect square trinomial plus a constant. The general form can be rewritten as a times x plus b over 2a squared, plus a constant term. To solve by completing the square, first ensure the coefficient of x squared is 1 by dividing the equation by a if needed. Then move the constant term to the right side. Next, take half of the coefficient of x, square it, and add this value to both sides. Rewrite the left side as a perfect square, take the square root of both sides, and finally solve for x. Let's work through an example: x squared plus 6x plus 5 equals zero. First, we rearrange to get x squared plus 6x equals negative 5. Half of 6 is 3, and 3 squared is 9. We add 9 to both sides: x squared plus 6x plus 9 equals negative 5 plus 9, which simplifies to x plus 3 squared equals 4. Taking the square root of both sides gives us x plus 3 equals plus or minus 2. Solving for x, we get x equals negative 3 plus or minus 2, which gives us our two solutions: x equals negative 1 or x equals negative 5.
Let's summarize what we've learned about solving quadratic equations. A quadratic equation has the form a x squared plus b x plus c equals zero, where a is not equal to zero. We've explored three methods to solve these equations. First, factoring, which works when the expression can be written as a product of linear factors. Second, the quadratic formula, which works for any quadratic equation and gives us the solutions directly. Third, completing the square, which transforms the quadratic into a perfect square trinomial plus a constant. This method is particularly useful for deriving the quadratic formula and converting to vertex form. Always remember to check your solutions by substituting them back into the original equation. Finally, the discriminant, b squared minus 4ac, tells us about the nature of the solutions: whether we have two distinct real solutions, one repeated real solution, or two complex solutions. By mastering these methods, you can solve any quadratic equation you encounter.