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 .
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 .
According to question ,we have
From equation ?. x2 = 729
? x = ± ?729 = ± 27
From equation ?. y = ?729
? y = 27
Hence , x ? y is required answer .
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 .
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 .
The given quadratic equation 3x2 + (k ? 1)x + 9 = 0
Putting x = 3 in above given eq. , we get
27 + 3(k ? 1) + 9 = 0
? 27 + 3k - 3 + 9 = 0
? 3k = ?33 ? k = ?11.
Hence , the value of k is ?11.
One root of the equation 3x2 - 10x + 3 = 0 is |
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. Find the other root. |
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The given expression x4 + 7x2 + 16 = ( x4 + 8x2 + 16 ) - x2
= ( x4 + 2 * 4 * x2 + 42 ) - x2
= ( x2 + 4 )2 - x2
= ( x2 + 4 + x ) ( x2 + 4 - x )
Hence , x4 + 7x2 + 16 = ( x2 + x + 4 ) ( x2 - x + 4 ).
Option C is correct answer .
As per the given above question , we can say that
The given quadratic equations are
x2 - 7x + 10 = 0 ? (x ? 5)(x ? 2) = 0 ? x = 5, 2
And x2 - 10x + 16 = 0 ? (x ? 8)(x ? 2) = 0 ? x = 8, 2
? Common root is 2.
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