If the root of x² + px + 12 = 0 are in the ratio 1:3 then what is the value of p ?

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Find the distance between the points ( l + m , n - p ) and ( l - m , n + p )

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If  \alpha , \beta are the roots of the equation x² + x√ \alpha +  \beta = 0 , then find  \alpha , and, \beta

2

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The Brainliest Answer!
2015-02-10T12:27:43+05:30

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 x^{2} + px + 12 = 0
so let the roots be 1 d and 3 d
We know that sum of the roots = -b/a
and product of the roots = c/a
1d+3d = - p / 1

1d * 3d = 12
 3d^{2} = 12
d = 2
So substituting, 1(2) + 3 (2) = - p
2+6 = - p
p = - 8 or +8 (since root can be both positive and negative)
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Distance between the points =  \sqrt{(x1-x2)^{2}+ (y1-y2)^{2}  }

= \sqrt{ (l+m-(l-m))^{2} +(n-p-(n+p)) ^{2} }

= \sqrt{ (2m)^{2} + (-2p) ^{2} }

= \sqrt{4m+4p}

=2 \sqrt{m+p} units
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 \alpha + \beta = -\frac{ \sqrt{\alpha } }{1}

 \alpha  \beta =  \frac{ \beta }{1}

 \alpha =1

sub alpha value,

1+ \beta = -  \sqrt{1}

1+ \beta =-1

 \beta  = - 2


1 5 1
2015-02-11T17:49:48+05:30
+ px + 12 = 0
so let the roots be 1 d and 3 d
We know that sum of the roots = -b/a
and product of the roots = c/a
1d+3d = - p / 1

1d * 3d = 12
= 12
d = 2
So substituting, 1(2) + 3 (2) = - p
2+6 = - p
p = - 8 or +8 (since root can be both positive and negative)
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Distance between the points = 

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sub alpha value,






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