Solving Quadratic Equations by Factoring Zero Factor Theorem
Solving Quadratic Equations by Factoring
Zero Factor Theorem Quadratic Equations • Can be written in the form ax 2 + bx + c = 0. • a, b and c are real numbers and a 0. • This is referred to as standard form. Zero Factor Theorem • If a and b are real numbers and ab = 0, then a = 0 or b = 0. • This theorem is very useful in solving quadratic equations.
Solving Quadratic Equations Steps for Solving a Quadratic Equation by Factoring 1) 2) 3) 4) 5) Write the equation in standard form. Factor the quadratic completely. Set each factor containing a variable equal to 0. Solve the resulting equations. Check each solution in the original equation.
Solving Quadratic Equations Example Solve x 2 – 5 x = 24. • First write the quadratic equation in standard form. x 2 – 5 x – 24 = 0 • Now we factor the quadratic using techniques from the previous sections. x 2 – 5 x – 24 = (x – 8)(x + 3) = 0 • We set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = – 3 Continued.
Solving Quadratic Equations Example Continued • Check both possible answers in the original equation. 82 – 5(8) = 64 – 40 = 24 true (– 3)2 – 5(– 3) = 9 – (– 15) = 24 true • So our solutions for x are 8 or – 3.
Solving Quadratic Equations Example Solve 4 x(8 x + 9) = 5 • First write the quadratic equation in standard form. 32 x 2 + 36 x = 5 32 x 2 + 36 x – 5 = 0 • Now we factor the quadratic. 32 x 2 + 36 x – 5 = (8 x – 1)(4 x + 5) = 0 • We set each factor equal to 0. 8 x – 1 = 0 or 4 x + 5 = 0 8 x = 1 or 4 x = – 5, which simplifies to x = or Continued.
Finding x-intercepts Recall earlier we found the x-intercept(s) of linear equations by letting y = 0 and solving for x. The same method works for x-intercepts in quadratic equations. Note: When the quadratic equation is written in standard form, the graph is a parabola opening up (when a > 0) or down (when a < 0), where a is the coefficient of the x 2 term. The intercepts will be where the parabola crosses the x-axis.
Finding x-intercepts Example Find the x-intercepts of the graph of f(x)= 4 x 2 + 11 x + 6. The equation is already written in standard form, so we let f(x) = 0, then factor the quadratic in x. 0 = 4 x 2 + 11 x + 6 = (4 x + 3)(x + 2) We set each factor equal to 0 and solve for x. 4 x + 3 = 0 or x + 2 = 0 4 x = – 3 or x = – 2 x = –¾ or x = – 2 So the x-intercepts are the points (–¾, 0) and (– 2, 0).
Quadratic Equations and Problem Solving
Strategy for Problem Solving General Strategy for Problem Solving 1) Understand the problem • Read and reread the problem • Choose a variable to represent the unknown • Construct a drawing, whenever possible • Propose a solution and check 2) Translate the problem into an equation 3) Solve the equation 4) Interpret the result • Check proposed solution in problem • State your conclusion
Finding an Unknown Number Example The product of two consecutive positive integers is 132. Find the two integers. 1. ) Understand Read and reread the problem. If we let x = one of the unknown positive integers, then x + 1 = the next consecutive positive integer. Continued
Finding an Unknown Number Example continued 2. ) Translate The product of two consecutive positive integers x • (x + 1) is 132 = 132 Continued
Finding an Unknown Number Example continued 3. ) Solve x(x + 1) = 132 x 2 + x – 132 = 0 (x + 12)(x – 11) = 0 x + 12 = 0 or x – 11 = 0 x = – 12 or x = 11 (Distributive property) (Write quadratic in standard form) (Factor quadratic polynomial) (Set factors equal to 0) (Solve each factor for x) Continued
Finding an Unknown Number Example continued 4. ) Interpret Check: Remember that x is suppose to represent a positive integer. So, although x = -12 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 11, then x + 1 = 12. The product of the two numbers is 11 · 12 = 132, our desired result. State: The two positive integers are 11 and 12.
The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. (leg a)2 + (leg b)2 = (hypotenuse)2
The Pythagorean Theorem Example Find the length of the shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg. 1. ) Understand Read and reread the problem. If we let x = the length of the shorter leg, then x + 10 = the length of the longer leg and 2 x – 10 = the length of the hypotenuse. Continued
The Pythagorean Theorem Example continued 2. ) Translate By the Pythagorean Theorem, (leg a)2 + (leg b)2 = (hypotenuse)2 x 2 + (x + 10)2 = (2 x – 10)2 3. ) Solve x 2 + (x + 10)2 = (2 x – 10)2 x 2 + 20 x + 100 = 4 x 2 – 40 x + 100 2 x 2 + 20 x + 100 = 4 x 2 – 40 x + 100 (multiply the binomials) (simplify left side) 0 = 2 x 2 – 60 x (subtract 2 x 2 + 20 x + 100 from both sides) 0 = 2 x(x – 30) (factor right side) x = 0 or x = 30 (set each factor = 0 and solve)Continued
The Pythagorean Theorem Example continued 4. ) Interpret Check: Remember that x is suppose to represent the length of the shorter side. So, although x = 0 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 30, then x + 10 = 40 and 2 x – 10 = 50. Since 302 + 402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem checks out. State: The length of the shorter leg is 30 miles. (Remember that is all we were asked for in this problem. )
- Slides: 18