Roots of a Quadratic

Let the roots be r and 2r+1 $(x-r)(x-2r-1)=0$ $=> x^{2}-(3m+1)x+2m^{2}+m=0$ From the equation, comparing coefficients of x we get $=> 3m+1=62/2=31$ i.e. $m=10$ and comparing coefficients of last term we get $k/2=2m^{2}+m=210 => k=420$

let the first root be "t" then by given condition the second root is "2t+1" now, sum of roots= t+2t+1 = -b/a = 62/2 so that t=10 product of roots t*(2t+1)= c/a = k/2 by putting value of t , k=420

let roots are x, 2x+1. 31=x+2x+1 x=10 K/2=10*21 k=420

Let one of the roots be $\alpha$

and let the other root be ${2}\alpha+{1}$

from the equation given, ${a}={2}$ ${b}={-62}$

and ${c}={k}$

sum of the roots equals $\alpha+{2}\alpha+{1}=\frac{-b}{a}=\frac{62}{2}$

${3}\alpha+{1}={31}$

$\alpha={10}$

product of the roots equals $\alpha({2}\alpha+{1})=\frac{c}{a}=\frac{k}{2}$

${2}(\alpha)^2+\alpha=\frac{k}{2}$

${k}={4}(\alpha)^2+{2}\alpha$

but $\alpha={10}$

thus, ${k}={4}({10})^2+{2}({10})$

${k}={400}+{20}$

Hence ${k}=\boxed{420}$

x1+x2= -b/a

x1 *x2= c/a

x1+2*(x2) +1 = -(-62)/2>>>>x1=10

x2=2*x2 +1>>>>> X2=21

10 *21= K/2

K= 420

two real roots m,n m+n= - (-62)/2 =31, m=2n+1, 2n+1+n=31, n=10, m=21,

product of two roots =m n =21 10=210=k/2, k=420.

the equation has 2 real roots which are x1 and x2 a=2 b=-62 c=k

x2=2x1+1

x1+x2=x1+2x1+1=-b/a=62/2=31

x1=10

x1 x2=x1 (2x1+1)=210=c/a=k/2

therefore k=210*2 =420