WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form a x + b = 0 , a ≠ 0 , and quadratic equations can be written in the general form a x 2 + b x + c = 0 , a ≠ 0 . We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of a x 2 + b x + c = 0 , b 2 − 4 a c , determines the number and type of solutions. We performed operations with complex numbers and used the imaginary unit i ( i = − 1 , where i 2 = − 1 ) to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x -intercepts. We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions. In Exercises 1-12, solve each equation. 3 x 4 − x 3 + 1 = 4 x 5 − 3 20
WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form a x + b = 0 , a ≠ 0 , and quadratic equations can be written in the general form a x 2 + b x + c = 0 , a ≠ 0 . We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of a x 2 + b x + c = 0 , b 2 − 4 a c , determines the number and type of solutions. We performed operations with complex numbers and used the imaginary unit i ( i = − 1 , where i 2 = − 1 ) to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x -intercepts. We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions. In Exercises 1-12, solve each equation. 3 x 4 − x 3 + 1 = 4 x 5 − 3 20
Solution Summary: The author calculates the solution set of the given equation as underset_x=3.
WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form
a
x
+
b
=
0
,
a
≠
0
, and quadratic equations can be written in the general form
a
x
2
+
b
x
+
c
=
0
,
a
≠
0
. We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of
a
x
2
+
b
x
+
c
=
0
,
b
2
−
4
a
c
, determines the number and type of solutions. We performed operations with complex numbers and used the imaginary unit
i
(
i
=
−
1
,
where
i
2
=
−
1
)
to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x-intercepts. We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions.
In Exercises 1-12, solve each equation.
3
x
4
−
x
3
+
1
=
4
x
5
−
3
20
Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity
+
Theorem: Let be a function from a topological
space (X,T) on to a non-empty set y then
is a quotient map iff
vesy if f(B) is closed in X then & is
>Y. ie Bclosed in
bp
closed in the quotient topology induced by f
iff (B) is closed in x-
التاريخ
Acy
الموضوع :
Theorem:- IP & and I are topological space
and fix sy is continuous
او
function and either
open or closed then the topology Cony is the
quatient topology p
proof:
Theorem: Lety have the quotient topology
induced by map f of X onto y.
The-x:
then an arbirary map g:y 7 is continuous
7.
iff gof: x > z is
"g of continuous
Continuous function
f
For the problem below, what are the possible solutions for x? Select all that apply.
2
x²+8x +11 = 0
x2+8x+16 =
(x+4)² = 5
1116
For the problem below, what are the possible solutions for x? Select all that apply.
x² + 12x - 62 =
0
x² + 12x + 36 = 62 + 36
(x+6)² = 98
Chapter 1 Solutions
Algebra & Trigonometry With Additional Material From College Algebra Essentials (custom Edition For Tidewater Community College)
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Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY