Consider a graph of quadratic equation y = ax² + bx + c as shown below where x₁ and x₂ are the roots of the equation ax² + bx + c = 0 , which of the following is/are correct
I) a - b - c <0 ii) bc<0 iii)b>0 iv) b and c have the same sign different from a
#Note
This question is multiple correct options
Answers
Topic :-
Quadratic Equation
Given :-
A graph of quadratic equation y = ax² + bx + c where x₁ and x₂ are roots of equation ax² + bx + c = 0.
To Find :-
Correct options among given options.
Solution :-
If mouth of graph opens upward then a > 0.
If mouth of graph opens downward then a < 0.
We can observe in the graph that its mouth open downwards. Hence, a < 0.
We can also observe that x₁ < 0 and x₂ > 0.
It is also clearly visible that |x₂| > |x₁| as root x₂ is much away from Y-axis as compared to root x₁.
Sum of roots of a quadratic equation equals to -b/a.
So, x₁ + x₂ = -b/a
As x₂ is positive and greater in magnitude than x₁. Hence, x₁ + x₂ > 0.
If x₁ + x₂ > 0 then -b/a will be also greater than 0.
-b/a > 0
'b' will have to be a positive number so that above inequality gets satisfied because a < 0. Hence, b > 0.
Product of roots of a quadratic equation equals to c/a.
So, x₁x₂ = c/a
As x₁ < 0 and x₂ > 0 hence their product will be always negative.
So, x₁x₂ < 0
which means c/a will be also negative.
c/a < 0
c will have to be positive as a < 0. Hence, c > 0.
We will check options now :
i) a - b - c < 0
As a < 0, b > 0 and c > 0, we are basically adding three negative numbers. Hence, their sum will be always a negative number.
Hence, this option is correct option.
ii) bc < 0
As b > 0 and c > 0, their product will be always a positive number. Hence, this option is wrong option.
iii) b > 0
It has been already checked above. As b > 0, this option is correct option.
iv) 'b' and 'c' have same sign and different from 'a'
b > 0 and c > 0 which means 'b' and 'c' have same sign.
a < 0 which means it has different sign from 'b' and 'c'.
Hence, this option is correct option.
Answer :-
Options i, iii and iv are correct options.