olve 2x2 + 3x + 5 = 0. Round solutions to the nearest hundredth. x ≈ −2.14 and x ≈ 0.64 x ≈ −0.64 and x ≈ 2.14 x ≈ −4.28 and x ≈ 1.28 No real solutions i got c am i correct?
Discriminant is b^2 - 4ac (This is the part that is inside the square root when you use the formula for finding roots to a quadratic equation). If discriminant is negative it means you have to take the square root of a negative number which implies there are no real solutions. Is the discriminant positive or negative here?
positive
or negative?
-31
If it is negative then no real solution exists for this equation.
so this is no solutions?
No REAL solutions. (But complex or imaginary solution does exist for this problem. But no REAL solution.)
ok thank youu
you are welcome.
Can you help me with another please?
@ranga
I can try.
ok
Identify the graph of a quadratic equation with two identical rational solutions.
The graph of which quadratic equation is shown below?
and that one
The roots or solutions of a quadratic equation on a graph is where the curve intersects the x axis: If the curve cuts the x axis at two points it means two different solutions. If the curve cuts the x axis at just one point it means two identical solutions. If the curve does not cut the x axis at all it means no real solutions So which one do you think is the answer?
its either c or d
d?
No. They are asking for the graph that shows two IDENTICAL rational solutions. It means the graph has to cut (or touch actually) at just ONE point.
OHHH its B
Yes.
What about the second one
What are the answer choices?
y = x^2 + 2x − 8 y = x^2 − 2x − 8 y = −x^2 − 2x + 8 y = −x^2 − 4x − 8
The general quadratic equation is: ax^2 + bx + c = 0 If a is positive, the graph will open upward. If a is negative, the graph will open downward. So right away you can eliminate two possibilities. Then notice that when x = 0, y = +8 in the graph. Which of the remaining two choices satisfy that condition?
the third one
yes.
thank you so much for your help and Happy Halloween
you are welcome. Happy Halloween to you too.
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