Zeroes of polynomial p(x)=2x^2-9-3x are :
1 answer:
Answer:
x = - , x = 3
Step-by-step explanation:
To find the zeros let p(x) = 0 , that is
2x² - 3x - 9 = 0
Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 2 × - 9 = - 18 and sum = - 3
The factors are - 6 and + 3
Use these factors to split the x- term
2x² - 6x + 3x - 9 = 0 ( factor the first/second and third/fourth terms)
2x(x - 3) + 3(x - 3) = 0 ← factor out (x - 3) from each term
(x - 3)(2x + 3) = 0
Equate each factor to zero and solve for x
2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = -
x - 3 = 0 ⇒ x = 3
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The dilation of the rectangle stretches it with regards to the distances from the central point
The option that gives the coordinates of the point <em>D</em> is the option;
D (0, 5)
Reason:
The known parameter;
Coordinates of the center of dilation is M(-3, -4)
Scale factor of dilation = 3
The coordinates of the vertex of the rectangle ABCD are;
A(-2, 2), B(3, 2), C(3, -1), and D(-2, -1)
Length MD =
Therefore, length of MD' = 3·√10
Slope of MD, is given as follows;
Therefore, y = 3·x
x² + y² = (3·√10)²
x² + (3·x)² = (3·√10)²
10·x² = 90
x = 3
y = 9
We note that the value of <em>x</em>, and <em>y</em>, are calculated as distance from the central point, <em>M</em>, therefore;
The coordinates of the point D' is (-3 + x, -4 + y)
∴ Coordinates of D' = (-3 + 3, -4 + 9) = (0, 5)
- Coordinates of D' = (0, 5)
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