Question

If a, b are the roots of quadratic equation $x^2 + px + q = 0$ and g, d are the roots of $x^2 + px - r = 0$, then (a - g). (a - d) is equal to:

• A. q+r
• B. q-r
• C.
  • (q+r)
• D. -(p+q+г)

Answer 1

Answer: (C)

-(q+r)

Answer 2

Let the given quadratic equations be:

1. $x^2 + px + q = 0$, with roots $a$ and $b$.
2. $x^2 + px - r = 0$, with roots $g$ and $d$.

From the properties of roots of a quadratic equation, for the second equation $x^2 + px - r = 0$, the roots are $g$ and $d$. This means that the polynomial $x^2 + px - r$ can be factored as $(x-g)(x-d)$. So, for any value of $x$, we have: $x^2 + px - r = (x-g)(x-d)$

We are asked to find the value of $(a - g) \cdot (a - d)$. This expression is the value of the polynomial $(x-g)(x-d)$ when $x = a$. Substituting $x = a$ into the equation $x^2 + px - r = (x-g)(x-d)$, we get: $a^2 + pa - r = (a-g)(a-d)$

Now, we use the information about the first equation. The roots of $x^2 + px + q = 0$ are $a$ and $b$. Since $a$ is a root of the equation $x^2 + px + q = 0$, it must satisfy the equation. So, substituting $x = a$ into the first equation, we get: $a^2 + pa + q = 0$

From this equation, we can express $a^2 + pa$ as: $a^2 + pa = -q$

Now, substitute this value of $a^2 + pa$ into the expression for $(a-g)(a-d)$: $(a-g)(a-d) = (a^2 + pa) - r$ $(a-g)(a-d) = (-q) - r$ $(a-g)(a-d) = -(q + r)$

Thus, the value of $(a - g) \cdot (a - d)$ is $-(q + r)$.