Question

If $x^3 - ax^2 + bx - a = 0$ has 3 real roots, then

• A. $b = 1$
• B. $b \neq 1$
• C. $a = 1$
• D. $a \neq 1$

Hint:

For cubic with all real roots, Vieta’s relations plus symmetry/root substitution often fix parameters quickly.

Answer 1

The Correct Option is A

Let the three real roots be $p, q, r$. From Vieta’s formulas: \[ p+q+r = a \] \[ pq + qr + rp = b \] \[ pqr = a \] Given $pqr = a$ and $p+q+r = a$, for three real roots equality of product and sum occurs when one root = 1 and the sum of other two equals their product (special symmetric case). Another way: factorize by grouping: \[ x^3 - ax^2 + bx - a = (x^2 + b)(x - a) + \dots \] Testing $x=1$ as a root: $1 - a + b - a = b - 2a + 1 = 0$. For all three real, $x=1$ must be a root, and discriminant conditions yield $b=1$. \[ \boxed{b=1} \]