Given :- Px2 + 4x + 1 = 0
Compare with Ax 2 + Bx + C = 0, we get
A = P, B = 4, C = 1
For real roots,
B2 - 4AC ? 0
? 16 - 4P ? 0
? 16 ? 4P
? P ? 16/4 ? P ? 4.
Hence , required answer is option C .
As per the given above question , we have
From equation ?. x2 - 24x + 144 = 0
? x2 ? 12x - 12x + 144 = 0
? x(x ? 12) ? 12(x ? 12) = 0
? (x ? 12)2 = 0
? x = 12
From equation ?. y2 ? 26y + 169 = 0
? y2 ? 13y ? 13y + 169 = 0
? y(y ? 13) ? 13(y ? 13) = 0
? (y ? 13)2 = 0
? y = 13
Hence, required answer will be x < y .
Given :- x2 - x - 2 = 0
The given equation is of the form
ax2 + bx + c = 0
Here , a = 1 , b = -1 , c = - 2
Now, D2 = b2 - 4ac = ( -1 )2 - 4 x 1 x ( - 2 )
D = ?9 = 3
So, roots are rational .
Hence, both the roots must be integers.
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