Question

The roots of the equation $b{x}^2+(b-c)x+(b-c-a)=0$ are real if those of $a{x}^2+2bx+b=0$ are imaginary.

• A. True
• B. False

Answer 1

The correct option is A True

Given that,

Roots of $a{x}^2+2bx+b=0$ are imaginary.

i.e., Discriminant $D<0$

$⟹b^2-4ac<0$

$⟹(2b{)}^2-4ab<0$

$⟹4b^2-4ab<0$

$⟹4ab-b^2>0$

Now, $b{x}^2+(b-c)x+(b-c-a)=0$

Here, $a=b,$

$b=(b-c),$ and

$c=(b-c-a)$

Discriminant $D=b^2-4ac$

$=(b-c{)}^2-4b(b-c-a)$

$=b^2+c^2-2bc-4b^2+4bc+4ab$

$=b^2+c^2+2bc+4ab-4b^2$

$=(b+c{)}^2+(4ab-4b^2)$

$\ge 0$

$\because (b+c{)}^2\ge 0$ and $4ab-4b^2>0$

so. the statement is true.