Question

If roots of equation $2x^2+bx+c=0;b,c\in R$, are real & distinct then the roots of equation $2{cx}^2+(b-4c)x+2c-b+1=0$ are

• A. imaginary
• B. equal
• C. real and distinct
• D. cant say

Answer 1

The correct option is C real and distinct

$1$st Equation is

$2x^2+bx+c=0$ $\to$ root are real distinct

$\therefore D>0$

$b^2-4ac>0$

$b^2-8c>0$ ........ (1)

$2$nd equation is

$2c{x}^2+(b-4c)x+2c-b+1=0$

To find the nature of roots, calculate $=b^2-4ac$

$D=(b-4c{)}^2-4(2c\left)\right(2c-b+1)$

$D=b^2-8c$

From (1)

It is +ve, so, roots are real and distinct.