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 + 3 ) 2 = 24
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 + 3 ) 2 = 24
Solution Summary: The author explains how to calculate the solution of the equation (x+3)2=24.
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
+
3
)
2
=
24
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
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Determine if the values of the variables listed are solutions of the system of equations.
2x - y = 4
3x+5y= - 6
x=1, y = 2; (1,-2)
Is (1, 2) a solution of the system of equations?
L
No
Yes
iew an example
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10. A sound wave's amplitude can be modeled by the function y = −7 sin ((x-1) + 4). Within the interval 0 < x < 12, when does the function have an amplitude
of 4? (Select all that apply.)
9.522 seconds
4.199 seconds
0.522 seconds
1.199 seconds
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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