solving quadratic equations by factoring. martin-gay, developmental mathematics 2 zero factor...
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Martin-Gay, Developmental Mathematics 3 Steps for Solving a Quadratic Equation by Factoring 1)Write the equation in standard form. 2)Factor the quadratic completely. 3)Set each factor containing a variable equal to 0. 4)Solve the resulting equations. 5)Check each solution in the original equation. Solving Quadratic EquationsTRANSCRIPT
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Solving Quadratic Equations by Factoring
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Martin-Gay, Developmental Mathematics 2
Zero Factor Theorem
Quadratic Equations• Can be written in the form ax2 + 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.
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Martin-Gay, Developmental Mathematics 3
Steps for Solving a Quadratic Equation by Factoring
1) Write the equation in standard form.2) Factor the quadratic completely.3) Set each factor containing a variable equal to 0.4) Solve the resulting equations.5) Check each solution in the original equation.
Solving Quadratic Equations
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Martin-Gay, Developmental Mathematics 4
Solve x2 – 5x = 24.• First write the quadratic equation in standard form.
x2 – 5x – 24 = 0• Now we factor the quadratic using techniques from
the previous sections. x2 – 5x – 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
Solving Quadratic Equations
Example
Continued.
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Martin-Gay, Developmental Mathematics 5
• 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.
Example Continued
Solving Quadratic Equations
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Martin-Gay, Developmental Mathematics 6
Solve 4x(8x + 9) = 5• First write the quadratic equation in standard form.
32x2 + 36x = 5 32x2 + 36x – 5 = 0
• Now we factor the quadratic using techniques from the previous sections.
32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0• We set each factor equal to 0.
8x – 1 = 0 or 4x + 5 = 0
Solving Quadratic Equations
Example
Continued.
8x = 1 or 4x = – 5, which simplifies to x = or5.418
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Martin-Gay, Developmental Mathematics 7
• Check both possible answers in the original equation.
1 1 14 8 9 4 1 9 4 (10) (10) 581
818 8 2
true
5 54 8 9 4 10 9 4 ( 1) ( 5)( 1) 545 54 44
true
• So our solutions for x are or .81
45
Example Continued
Solving Quadratic Equations
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Martin-Gay, Developmental Mathematics 8
Previously, we found the x-intercept 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 x2 term.The intercepts will be where the parabola crosses the x-axis.
Finding x-intercepts
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Martin-Gay, Developmental Mathematics 9
Find the x-intercepts of the graph of y = 4x2 + 11x + 6.The equation is already written in standard form, so we let y = 0, then factor the quadratic in x.0 = 4x2 + 11x + 6 = (4x + 3)(x + 2)We set each factor equal to 0 and solve for x.4x + 3 = 0 or x + 2 = 04x = –3 or x = –2x = –¾ or x = –2So the x-intercepts are the points (–¾, 0) and (–2, 0).
Finding x-intercepts
Example