According to question , we have
The roots ?, ? of the equation : x2 ? 6x + k = 0
On comparing with ax2 + bx + c = 0 , we get a = 1 , b = - 6 , c = k
Now , ? + ? = ( - b/a ) = 6
? + ? = 6 and 3? + 2? = 20
On solving ,
? ? = 4, ? = 2
Product of the roots = k
So, k = ?? = 4 × 2 = 8.
Hence , required answer will be option A .
According to question ,
?. x2 ? 20x + 91= 0
? x2 ? 13x ? 7x + 91 = 0
? x(x ? 13) ? 7(x ? 13) = 0
? (x ? 7)(x ? 13) = 0
? x = 13, 7
?. y2 ? 32y + 247= 0
? y2 ? 19y ? 13y + 247 = 0
? y(y ? 19) ? 13(y ? 19) = 0
? (y ? 13)(y ? 19) = 0
? y = 13, 19
Hence, x ? y is required answer .
As per the given above question , we know that
Let, ?, ? be the roots of the equation x2 + kx + 12 = 0
? ? + ? = ?k and ?? = 12
Now ( ? - ? )2 = ( ? + ? )2 - 4??
( 1 )2 = k2 - 4(12)
? k2 = 49 ? k = ± 7.
find ? if ?2 + ?2 = |
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a and b are the roots of the equation
x2 - 6x + 6 = 0
? a + b = -B/A = 6 and ab = C/A = 6
We know that,
a2 + b2 = (a + b)2 - 2ab
= (6)2 - 2 x 6 = 36 - 12 = 24
? 2(a2 + b2) = 2 x 24 = 48
Given, ? and ? are the roots of the equation ax2 + bx + c = 0
? Sum of two roots = -b/a
? + ? = -b/a
and product of two roots = c/a
? ? = c/a
? ?2/? + ?2/? = ( ?3 + ?3) / ? ?
= [(? + ?) 3 - 3 ? ? (? + ?)] / ??
[? (a + b)3 = a3 + b3 + 3ab(a + b) ]
= (-b/a)3 - [3c/a(-b/a)/c/a = -b3/a3 + 3bc/a2]/(c/a)
= 3abc - b3/a2c
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