Question

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 $\alpha , and, \beta$

Answer 1

$x^{2}$ + 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,

$1+ \beta = - \sqrt{1}$

$1+ \beta =-1$

$\beta = - 2$

Answer 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
= 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,