According to question ,we have
From equation ?. x2 ? 4 = 0
? x = ± 2,
From equation ?. y2 + 6y + 9 = 0
? ( y + 3 )2 = 0
? ( y + 3 )( y + 3 ) = 0
? y + 3 = 0 ? y = -3 , -3
Clearly , x > y is required answer .
Let, the two natural consecutive odd numbers be n and (n + 2)
Now, according to the question,
? n2 + ( n + 2 )2 = 394
? n2 + n2 + 4 + 4n = 394
? 2n2 + 4n - 390= 0
? n2 + 2n - 195 = 0
? n2 + 15n - 13n - 195 = 0
? n( n + 15 ) - 13( n + 15 ) = 0
? ( n + 15 ) ( n - 13 ) = 0
? n = 13 and n ? - 15
? The two natural consecutive odd numbers be 13 and (13 + 2) = 15 .
? the sum of the numbers = 13 + 15 = 28
Quicker Approach:
By mental operation, 132 + 152 = 169 + 225 = 394
? Required sum = 13 + 15 = 28
According to question ,we can say that
Let , x = ?30 + ?30 + ?30 +........
On squaring both sides, we have
x2 = 30 + ?30 + ?30 + ?30 +........
? x2 = 30 + x ? x2 - x - 30 = 0
? x2 - 6x + 5x - 30 = 0
? x( x - 6 ) + 5( x - 6 ) = 0
? ( x - 6 ) ( x + 5 ) = 0
? x = 6 because x ? - 5
Hence , required answer is 6 .
Given that :- The factors of a2 + 4b2 + 4b ? 4ab ? 2a ? 8
= a2 + 4b2 ? 4ab ? 2a + 4b ? 8
= (a ? 2b) 2 ? 2(a ? 2b) ? 8
Let, (a ? 2b) = x
? The given expression = x2 ? 2x ? 8
= x2 ? 4x + 2x ? 8
= x(x ? 4) + 2(x ? 4)
= (x ? 4) (x + 2)
Putting the value of x , we get
? a2 + 4b2 + 4b ? 4ab ? 2a ? 8 = (a ? 2b ? 4) (a ? 2b + 2)
Therefore , required answer will be (a ? 2b ? 4) (a ? 2b + 2) .
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According to question ,we have
From equation ?. x2 = 729
? x = ± ?729 = ± 27
From equation ?. y = ?729
? y = 27
Hence , x ? y is required answer .
As per the given above equations , we have
From equation ?. x2 - 7x + 12 = 0
? x2 - 4x - 3x + 12 = 0
? x( x - 4) - 3( x - 4) = 0
? ( x - 4) ( x - 3) = 0
? x = 4 or, 3
?. y2 - 12y + 32 = 0
? y2 - 8y - 4y + 32 = 0
? y( y - 8) - 4( y - 8) = 0
? ( y - 8) ( y - 4) = 0
? y = 4 or, 8
Thus , x ? y is required answer .
Given that :- Given that :- ?. 5x + 2y = 31 .....( 1 )
?. 3x + 7y = 36 .....( 2 )
Solving these two linear equations, we get x = 5, y = 3 .
? x > y is correct answer .
According to question , we can say that
The equation x2 ? px + q = 0, p, q ? R has no real root
On comparing with quadratic eq. Ax2 + Bx + C = 0 , we get
? A =1, B= - p, C = q
Now ,we have B2 < 4AC
? ( - P )2 < 4 x 1 x q
? p2 < 4q.
Hence , option B is correct answer .
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