In a quadratic equation
q(x) = ax^2 + bx + c
The discriminant is = b^2 - 4ac
We have that discriminant = 3
If
b^2 - 4ac > 0, then the roots are real.
If
b^2 - 4ac < 0 then the roots are imaginary
<span>In
this problem b^2 - 4ac > 0 3 > 0 </span>
then
the two roots must be real
Answer:
Inequalities are,
y ≥ 4x + 2
y ≥ 2
Step-by-step explanation:
Solid yellow line of the graph attached passes through two points (0, -2) and (1, 2).
Let the equation of this line is,
y = mx + b
Slope of the line = 
m = 
m = 4
Y-intercept 'b' = -2
Equation of the line will be,
y = 4x - 2
Since shaded area is on the left side of this solid line so the inequality representing this region will be,
y ≥ 4x - 2
Another line is a solid blue line parallel to the x-axis.
Shaded region (blue) above the line will be represented by,
y ≥ 2
Therefore, the common shaded area of these inequalities will be the solution of the given inequalities.
Answer:
d one i think
hope it helps
the sum of triangle is 180 degree
Answer:
p = 8
Step-by-step explanation:
Let one root of the eqn. be alpha . Other root is 1/alpha .
We know that product of both roots of an quadratic eqn. is c/a where "c" is the co-efficient of the constant & "a" is the co-efficient of x^2.
Here "c" is p-4 & "a" is 4. And the product of roots is 1 ( ∵ prdouct of a number and its reciprocal is 1 )
