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 x ( 2 x − 3 ) = − 4
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 x ( 2 x − 3 ) = − 4
Solution Summary: The author explains how to calculate the solution of the quadratic equation x(2x-3)=-4.
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
x
(
2
x
−
3
)
=
−
4
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
Chapter 4 Quiz 2 As always, show your work. 1) FindΘgivencscΘ=1.045.
2) Find Θ given sec Θ = 4.213.
3) Find Θ given cot Θ = 0.579. Solve the following three right triangles.
B
21.0
34.6° ca
52.5
4)c
26°
5)
A
b
6) B 84.0 a
42°
b
Q1: A: Let M and N be two subspace of finite dimension linear space X, show that if M = N
then dim M = dim N but the converse need not to be true.
B: Let A and B two balanced subsets of a linear space X, show that whether An B and
AUB are balanced sets or nor.
Q2: Answer only two
A:Let M be a subset of a linear space X, show that M is a hyperplane of X iff there exists
ƒ€ X'/{0} and a € F such that M = (x = x/f&x) = x}.
fe
B:Show that every two norms on finite dimension linear space are equivalent
C: Let f be a linear function from a normed space X in to a normed space Y, show that
continuous at x, E X iff for any sequence (x) in X converge to Xo then the sequence
(f(x)) converge to (f(x)) in Y.
Q3: A:Let M be a closed subspace of a normed space X, constract a linear space X/M as
normed space
B: Let A be a finite dimension subspace of a Banach space X, show that A is closed.
C: Show that every finite dimension normed space is Banach space.
• Plane II is spanned by the vectors:
P12
P2 = 1
• Subspace W is spanned by the vectors:
W₁ =
-- () ·
2
1
W2 =
0
Elementary Linear Algebra (Classic Version) (2nd Edition) (Pearson Modern Classics for Advanced Mathematics Series)
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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