The error made by zacharias in solving the quadratic equation 0 = -2x² + 5x - 3 using quadratic formula is; the 2 in the numerator should be –2.
<h3>Quadratic equation</h3>
There are four methods of solving quadratic equation. Namely;
- Factorization method
- Completing the square method
- Graphical method
- Formula method
0 = -2x² + 5x - 3
x = -b ± √b² - 4ac / 2a
where,
x = -b ± √b² - 4ac / 2a
x = -5 ± √5² - 4(-2)(-3) / 2(-2)
= -5 ± √25 - (24) / -2
= -5 ± √1 / -2
= -5/2 ± 1/2
= -5/2 - 1/2 or x = -5/2 + 1/2
= -5-1 / 2 or -5+1/ 2
x = -6/2 or -4/2
x = -3 or -2
Therefore, the solution to the quadratic equation is x = -3 or -2
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Answer:
12 x 6 = 72, 72 divided by 2 ( x 1/2) = 36 yd^2
Formula of a triangle:
BH x 1/2
In this case 12 is your base and 6 is your height.
Answer:
84
Step-by-step explanation:
The interquartile range is obtained using the relation :
Third quartile (Q3). - First quartile (Q1)
From. The boxplot :
Q3 = 96
Q1 = 12
Interquartile range (IQR) = Q3 - Q1 = 96 - 12 = 84
Answer:
<h2>A. (0,1)</h2>
Step-by-step explanation:
The question lacks the e=required option. Find the complete question below with options.
Which of the following points does not belong to the quadratic function
f(x) = 1-x²?
a.(0,1) b.(1,0) c.(-1,0)
Let f(x) = 0
The equation becomes 1-x² = 0
Solving 1-x² = 0 for x;
subtract 1 from both sides;
1-x²-1 = 0-1
-x² = -1
multiply both sides by minus sign
-(-x²) = -(-1)
x² = 1
take square root of both sides;
√x² = ±√1
x = ±1
x = 1 and x = -1
when x = 1
f(x) = y = 1-1²
y = 1-1
y = 0
when x = -1
f(x) = y = 1-(-1)²
y = 1-1
y = 0
Hence the coordinate of the function f(x) = 1-x² are (±1, 0) i.e (1, 0) and (-1, 0). The point that does not belong to the quadratic function is (0, 1)
Answer:
The point will come in III quadrant
Step-by-step explanation:
The graph is as follows:
