If the root of x² + px + 12 = 0 are in the ratio 1:3 then what is the value of p ?=====================================================Find the distance between the points ( l + m , n - p ) and ( l - m , n + p )===================================================If are the roots of the equation x² + x√ + = 0 , then find

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

1.	If the root of x² + px + 12 = 0 are in the ratio 1:3 then what is the value of p ?=====================================================Find the distance between the points ( l + m , n - p ) and ( l - m , n + p )===================================================If are the roots of the equation x² + x√ + = 0 , then find
Answer» $x^{2}$ + 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) ================================================== $\sqrt{(x1-x2)^{2}+ (y1-y2)^{2} }$ DISTANCE between the points = $\sqrt{ (l+m-(l-m))^{2} +(n-p-(n+p)) ^{2} }$ = $\sqrt{ (2m)^{2} + (-2p) ^{2} }$ = $\sqrt{4m+4p}$ = $2 \sqrt{m+p}$ = units ===================================================== $\alpha + \beta = -\frac{ \sqrt{\alpha } }{1}$ $\alpha \beta = \frac{ \beta }{1}$ $\alpha =1$ sub alpha value,** $1+ \beta = - \sqrt{1}$ $1+ \beta =-1$ $\beta = - 2$