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pl. provide solutions

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Grade:12th pass

5 Answers

Yash Jain
55 Points
8 years ago
For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b2 - ac 2 - 4ac
Thus, roots are imaginary.
We have, 4a+4b+c2 - 2b(-2) + c
The required expression is a + 2b + c or a(-1)2 – 2b(-1) + c or f(-1)
Now understand the question by graphical method. Since the roots are imaginary, so the graph of f(x) is not going to intersect the x-axis. Also, f(-2)
Yash Jain
55 Points
8 years ago
For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b2 - ac 2 - 4ac 
Thus, roots are imaginary.
We have, 4a+4b+c2 → a(-2)2 - 2b(-2) + c → f(-2)
The required expression is a + 2b + c or a(-1)2 – 2b(-1) + c or f(-1)
Now understand the question by graphical method. Since the roots are imaginary, so the graph of f(x) is not going to intersect the x-axis. Also, f(-2)
Yash Jain
55 Points
8 years ago
For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b2 - ac 2 - 4ac 
Thus, roots are imaginary.
We have, 4a+4b+c2 → a(-2)2 - 2b(-2) + c → f(-2)
The required expression is a + 2b + c or a(-1)2 – 2b(-1) + c or f(-1)
Now understand the question by graphical method. Since the roots are imaginary, so the graph of f(x) is not going to intersect the x-axis. Also, f(-2)
Yash Jain
55 Points
8 years ago
For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b2 - ac 2 - 4ac 
Thus, roots are imaginary.
We have, 4a+4b+c2 → a(-2)2 - 2b(-2) + c → f(-2)
The required expression is a + 2b + c or a(-1)2 – 2b(-1) + c or f(-1)
Now understand the question by graphical method. Since the roots are imaginary, so the graph of f(x) is not going to intersect the x-axis. Also, f(-2)
Yash Jain
55 Points
8 years ago
Also, f(-2)

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