Flag Algebra> pl. provide solutions...
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pl. provide solutions

Y RAJYALAKSHMI , 10 Years ago
Grade 12th pass
anser 5 Answers
Yash Jain

Last Activity: 10 Years ago

For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b- ac - 4ac
Thus, roots are imaginary.
We have, 4a+4b+c- 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

Last Activity: 10 Years ago

For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b- ac - 4ac 
Thus, roots are imaginary.
We have, 4a+4b+c2 → a(-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

Last Activity: 10 Years ago

For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b- ac - 4ac 
Thus, roots are imaginary.
We have, 4a+4b+c→ a(-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

Last Activity: 10 Years ago

For your first question...
Given quadratic : ax2 – 2bx + c = 0
Now, b- ac - 4ac 
Thus, roots are imaginary.
We have, 4a+4b+c→ a(-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

Last Activity: 10 Years ago

Also, f(-2)

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