Example 1 utilized the Quadratic Formula . Verify that x = − b + b 2 − 4 a c 2 a is a solution of the equation a x 2 + b x + c = 0 . HINT: Substitute the fraction for x in a x 2 + b x + c and simplify.
Example 1 utilized the Quadratic Formula . Verify that x = − b + b 2 − 4 a c 2 a is a solution of the equation a x 2 + b x + c = 0 . HINT: Substitute the fraction for x in a x 2 + b x + c and simplify.
Solution Summary: The author explains that the given value of x is the solution for the quadratic equation.
Example 1 utilized the Quadratic Formula. Verify that
x
=
−
b
+
b
2
−
4
a
c
2
a
is a solution of the equation
a
x
2
+
b
x
+
c
=
0
.
HINT: Substitute the fraction for
x
in
a
x
2
+
b
x
+
c
and simplify.
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
Explain the key points and reasons for 12.8.2 (1) and 12.8.2 (2)
Q1:
A slider in a machine moves along a fixed straight rod. Its
distance x cm along the rod is given below for various values of the time. Find the
velocity and acceleration of the slider when t = 0.3 seconds.
t(seconds)
x(cm)
0 0.1 0.2 0.3 0.4 0.5 0.6
30.13 31.62 32.87 33.64 33.95 33.81 33.24
Q2:
Using the Runge-Kutta method of fourth order, solve for y atr = 1.2,
From
dy_2xy +et
=
dx x²+xc*
Take h=0.2.
given x = 1, y = 0
Q3:Approximate the solution of the following equation
using finite difference method.
ly -(1-y=
y = x), y(1) = 2 and y(3) = −1
On the interval (1≤x≤3).(taking h=0.5).
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