If b

Question

If b < 0, then the roots x1 and x2 of the equation 2 x2+6 x+b=0, satisfy the condition ((x1/x2))+((x2/x1)) < K, where K is equal to (A) 2 (B) -2 (C) 0 (D) 4

Tardigrade Admin · Accepted Answer

The discriminant of the quadratic equation 2 x2+6 x+b=0 is given by D=36-8 b>0. Therefore, the given equation has real roots. we have, (x1/x2)=(x2/x1)=(x12+x22/x1 ⋅ x2)=((x1+x2)2-2 x1 x2/x1 ⋅ x2) =((-3)2-2(b / 2)/(b / 2))=(18/b)-2<-2 [∵ b<0]

Q. If b<0, then the roots x1​ and x2​ of the equation 2x2+6x+b=0, satisfy the condition (x2​x1​​)+(x1​x2​​)<K, where K is equal to

2

-2

0

4

The discriminant of the quadratic equation 2x2+6x+b=0 is given by D=36−8b>0. Therefore, the given equation has real roots. we have, x2​x1​​=x1​x2​​=x1​⋅x2​x12​+x22​​=x1​⋅x2​(x1​+x2​)2−2x1​x2​​ =(b/2)(−3)2−2(b/2)​=b18​−2<−2 [∵b<0]

Q. If $b < 0$ , then the roots $x_{1}$ and $x_{2}$ of the equation $2 x^{2} + 6 x + b = 0$ , satisfy the condition $(\frac{x _{1}}{x _{2}}) + (\frac{x _{2}}{x _{1}}) < K$ , where $K$ is equal to