For
,
, all the more
, therefore no solution.
For
, we have
. Obviously that' false.
For
both
and
, where
, belong to the range
, then
belong to the range
.
Those ranges don't have a common part, therefore, again, no solution.
So, the equation
doesn't have a solution.
Answer:
If point C lies between two points A and B such that AC=BC, then point C is the bisector of A and B, that means it is at right the centre.
Answer:
Two times at (-1,0) and (2.5,0)
Step-by-step explanation:
When the graph intersects or touches x-axis, y is equal to 0
so y = -2x^2 + 3x + 5
=> 0 = -2x^2 + 3x + 5
The formula to solve a quadratic equation of the form ax^2 + bx + c = 0 is equal to x = [-b +/-√(b^2 - 4ac)]/2a
so a = -2
b = 3
c = 5
substitute in the formula
x = [-3 +/- √(3^2 - 4x-2x5)]/2(-2)
x = [-3 +/- √(9 + 40)]/(-4)
x = [-3 +/- 7]/(-4)
x1 = (-3 + 7)/(-4) = 4/-4 = -1
x2 = (-3 - 7)/(-4) = -10/-4 = 5/2 = 2.5
so the graph has two x-intercepts (-1,0) and (2.5,0), therefore it intersects x-axis two times
Answer:
y=1/6 x
Step-by-step explanation:
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