ary number, equal or two dis- ned by the value of b2- 4ac. Learning Task 1. Complete the table given below and find the relations among constants a, b and c in the quadratic equation standard form. Equation b. Roots a. C. b2 - 4ac X1 X2 x2 + 4x + 3-0 x2 - 5x + 4 0 x2 - 49 -0 4x2 - 25 - 0 2x2 + 7x + 3= 0

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter1: Equations And Graphs
Section: Chapter Questions
Problem 15CC
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tinct roots. It can be determined by the value of b2 -4ac.
uadratic equation can be imaginary number, equal or two dis-
Learning Task 1. Complete the table given below and find the relations among
constants a, b and c in the quadratic equation standard form.
Equation
Roots
a
b2 - 4ac
X1
X2
x2 + 4x + 3 = 0
x2 -5x + 4 0
x2-49=0
4x2 - 25 = 0
2x2 + 7x + 3 = 0
%3D
D
The expression b2 - 4ac, is the quadratic equation's discriminant. The
discriminant determines the nature of the roots of a quadratic equation.
If
b2 - 4ac =0, then the roots are real and equal
Transcribed Image Text:tinct roots. It can be determined by the value of b2 -4ac. uadratic equation can be imaginary number, equal or two dis- Learning Task 1. Complete the table given below and find the relations among constants a, b and c in the quadratic equation standard form. Equation Roots a b2 - 4ac X1 X2 x2 + 4x + 3 = 0 x2 -5x + 4 0 x2-49=0 4x2 - 25 = 0 2x2 + 7x + 3 = 0 %3D D The expression b2 - 4ac, is the quadratic equation's discriminant. The discriminant determines the nature of the roots of a quadratic equation. If b2 - 4ac =0, then the roots are real and equal
part of work done by Odette in a minute
1-15
10
Equation:
1-15
10
10
(t- 15) +t
t(1- 15)
1
%3D
Find the LCD, Then add:
t-15
10
2t -15
+ 10(2t – 15) = t2-15
2-15t
10
20t - 150 = t2-5 +t- 20t + 145 = 0
You may solve the equation by any method applicable.
Learning Task 2.
A. Translate the following verbal sentences to mathematical sentence. Thee
press into quadratic equations in terms of "x".
Given
Quadratic Equations
1. The length of a wooden frame is 1 foot
longer than its width and its area is equal
to 12 ft2.
2. The length of the floor is 8 m longer than
its width and there is 20 square meters.
3. The length of a plywood is 0.9 m more
than its width and its area is 0.36 m2.
4. The area of rectangle whose length is six
less than twice its width is thirty-six.
5. The width of a rectangular plot is 5 m
less than its length and its area is 84 m2.
B. Solve for x:
%3D
x+5
10
PIVOT 4A CALABARZON
20
117
Transcribed Image Text:part of work done by Odette in a minute 1-15 10 Equation: 1-15 10 10 (t- 15) +t t(1- 15) 1 %3D Find the LCD, Then add: t-15 10 2t -15 + 10(2t – 15) = t2-15 2-15t 10 20t - 150 = t2-5 +t- 20t + 145 = 0 You may solve the equation by any method applicable. Learning Task 2. A. Translate the following verbal sentences to mathematical sentence. Thee press into quadratic equations in terms of "x". Given Quadratic Equations 1. The length of a wooden frame is 1 foot longer than its width and its area is equal to 12 ft2. 2. The length of the floor is 8 m longer than its width and there is 20 square meters. 3. The length of a plywood is 0.9 m more than its width and its area is 0.36 m2. 4. The area of rectangle whose length is six less than twice its width is thirty-six. 5. The width of a rectangular plot is 5 m less than its length and its area is 84 m2. B. Solve for x: %3D x+5 10 PIVOT 4A CALABARZON 20 117
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