Solving Quadratic Equations Lesson 9-3

Standard form of a quadratic function A quadratic equation is a equation that can be written in the form 𝒂 𝒙 𝟐 +𝒃𝒙+𝒄=𝟎 Where 𝑎≠0. This form is called the standard form of a quadratic equation

Quadratic equation can be solved by a variety of methods Graphing Squared Root property Factoring Completing the square Quadratic formula

Solutions of a Quadratic Equation A quadratic equation can have two, one or no real number solutions The solutions of a quadratic equation are the x- intercepts of the related function The solutions of a quadratic equation are often called Roots of the equation or Zeros of the function

Steps in Solving Quadratic Equations If the equation is in the form (ax+b)2 = c, use the square root property to solve. If not solved in step 1, write the equation in standard form. Try to solve by factoring. If you haven’t solved it yet, use the quadratic formula.

are the x-intercepts 2 and -2 Solving Quadratic Equation by graphing Graph the parabola by: y x Finding the equation of the axis of symmetry The solution of 𝑥 2 −4=0 are the x-intercepts 2 and -2 𝑥= −𝑏 2𝑎 x= 0 2(1) 𝑥=0 Finding the vertex (0) 2 −4=−4 Vertex = (0,-4) Making a table using the x-values around the axis of symmetry 2 -2 x 𝒙 𝟐 −𝟒 y 2 (𝟐) 𝟐 −𝟒 -2 (−𝟐) 𝟐 −𝟒 Graphing each point on a coordinate plane The roots of the equation or solution are the x-intercept or zeros of the related quadratic function

Examples of solving by graphing What are the solution of each equation? Use a graph of the related function 𝑥 2 −1=0 𝑥 2 =0 𝑥 2 +1=0 There are two solutions 1 and -1 There is one solutions There is no real number solution

Solving Quadratic Equation by using the square root Square Root Property If b is a real number and a2 = b, then You can solve equations of the form 𝒙 𝟐 =𝒌 by finding the square root of each side. For example 𝑥 2 =81 𝑥=± 81 𝑥=±9

Solving Quadratic Equation by using the square root Example Solve: 𝐵) 3𝑥 2 −75=0 A) (y – 3)2 = 4 3𝑥 2 =75 𝑥 2 = 75 3 𝑦−3=±2 𝑥 2 = 25 𝑦=3±2 𝑦=1 𝑜𝑟 𝑦=5 𝑥=± 25 𝑥=±5

Solving Quadratic Equation by using Quadratic formula This method will work to solve ALL quadratic equations For many equations it takes longer than some of the other methods.

Solving Quadratic Equation by using Quadratic formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions. 𝒙= −𝒃± 𝒃 𝟐 −𝟒𝒂𝒄 𝟐𝒂 When getting your solution, if the radicand in the quadratic formula is not a perfect square, you can use a calculator to approximate the solution

Solve x2 + x – = 0 by the quadratic formula. Example Solve x2 + x – = 0 by the quadratic formula. x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20 = −8+12 2 or −8−12 2 = −20 2 𝑜𝑟 4 2 𝑥=−10 𝑜𝑟 𝑥=2

So there is no real solution. Example Solve x(x + 6) = 30 by the quadratic formula. (Simplify the polynomial and write it on standard form) x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution. Square root can’t be negative.

Solve 2x = x2 - 8. x2 – 2x – 8 = 0 Let a = 1, b = -2, c = -8 Example write the equation on standard form x2 – 2x – 8 = 0 Let a = 1, b = -2, c = -8 𝑥= 2+6 2 𝑜𝑟 𝑥= 2−6 2 𝑥=4 𝑜𝑟 𝑥=−2

The Discriminat Before you solve a quadratic equation you can determine how many real-number solutions it has by using discriminant. The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 Because the discriminant is negative, the equation has no-real number solution.

Because the discriminant is positive, the equation has two solution. Example Use the discriminant to determine the number and type of solutions for the following equation. 6 𝑥 2 −5𝑥=7 6 𝑥 2 −5𝑥−7=0 a = 6, b = –5, and c = -7 b2 – 4ac = (–6)2 – 4(6)(-7) = 36 +196 = 232 Because the discriminant is positive, the equation has two solution.