Question

If the roots of $ax^2 + 2bx + c = 0$ be possible and different, then the roots of

$(a + c)(ax^2 + 2bx + c) = 2(ac - b^2) (x^2 + 1)$ will be impossible, and vice versa.

Answer 1

The statement is proven

Answer 2

Let the first quadratic equation be $P(x) = ax^2 + 2bx + c = 0$. The roots of this equation are possible and different if its discriminant is positive. The discriminant is $D_1 = (2b)^2 - 4ac = 4b^2 - 4ac = 4(b^2 - ac)$.

The condition for possible and different roots is $D_1 > 0$, which means $4(b^2 - ac) > 0$, or $b^2 - ac > 0$.

Let the second equation be $(a + c)(ax^2 + 2bx + c) = 2(ac - b^2) (x^2 + 1)$.

We can rearrange this equation into the standard quadratic form $Ax^2 + Bx + C = 0$.

Expanding the equation:

$(a + c)ax^2 + (a + c)2bx + (a + c)c = 2(ac - b^2)x^2 + 2(ac - b^2)$

$a(a + c)x^2 + 2b(a + c)x + c(a + c) = 2(ac - b^2)x^2 + 2(ac - b^2)$

Moving all terms to the left side:

$[a(a + c) - 2(ac - b^2)]x^2 + [2b(a + c)]x + [c(a + c) - 2(ac - b^2)] = 0$

$[a^2 + ac - 2ac + 2b^2]x^2 + 2b(a + c)x + [ac + c^2 - 2ac + 2b^2] = 0$

$[a^2 - ac + 2b^2]x^2 + [2b(a + c)]x + [c^2 - ac + 2b^2] = 0$

Let the coefficients of the second equation be $A = a^2 - ac + 2b^2$, $B = 2b(a + c)$, and $C = c^2 - ac + 2b^2$.

The roots of the second equation are impossible (non-real) if its discriminant $D_2 = B^2 - 4AC$ is negative.

Let's calculate $D_2$:

$D_2 = [2b(a + c)]^2 - 4(a^2 - ac + 2b^2)(c^2 - ac + 2b^2)$

$D_2 = 4b^2(a + c)^2 - 4(a^2 - ac + 2b^2)(c^2 - ac + 2b^2)$

Let $k = b^2 - ac$. Then $b^2 = ac + k$. The condition for the first equation's roots to be possible and different is $k > 0$.

Substitute $b^2 = ac + k$ into the expression for $D_2$:

$A = a^2 - ac + 2(ac + k) = a^2 + ac + 2k$

$C = c^2 - ac + 2(ac + k) = c^2 + ac + 2k$

$B = 2b(a + c)$

$D_2 = 4(ac + k)(a + c)^2 - 4(a^2 + ac + 2k)(c^2 + ac + 2k)$

$D_2 = 4(ac + k)(a^2 + 2ac + c^2) - 4(a^2c^2 + a^3c + 2a^2k + ac^3 + a^2c^2 + 2ack + 2kc^2 + 2akc + 4k^2)$

$D_2 = 4(a^3c + 2a^2c^2 + ac^3 + a^2k + 2ack + c^2k) - 4(2a^2c^2 + a^3c + ac^3 + 2a^2k + 4ack + 2kc^2 + 4k^2)$

$D_2 = 4[a^3c + 2a^2c^2 + ac^3 + a^2k + 2ack + c^2k - (2a^2c^2 + a^3c + ac^3 + 2a^2k + 4ack + 2kc^2 + 4k^2)]$

$D_2 = 4[a^3c - a^3c + 2a^2c^2 - 2a^2c^2 + ac^3 - ac^3 + a^2k - 2a^2k + 2ack - 4ack + c^2k - 2kc^2 - 4k^2]$

$D_2 = 4[-a^2k - 2ack - c^2k - 4k^2]$

$D_2 = 4[-k(a^2 + 2ac + c^2) - 4k^2]$

$D_2 = 4[-k(a + c)^2 - 4k^2]$

$D_2 = -4k[(a + c)^2 + 4k]$

Part 1: If the roots of $ax^2 + 2bx + c = 0$ are possible and different, then $b^2 - ac > 0$.

Let $k = b^2 - ac$. So $k > 0$.

The discriminant of the second equation is $D_2 = -4k[(a + c)^2 + 4k]$.

Since $k > 0$, we have $4k > 0$.

Also, $(a + c)^2 \ge 0$ for real $a$ and $c$.

Therefore, $(a + c)^2 + 4k > 0$.

Since $k > 0$ and $(a + c)^2 + 4k > 0$, their product $k[(a + c)^2 + 4k]$ is positive.

Thus, $D_2 = -4 \times (\text{positive number}) = \text{negative number}$.

$D_2 < 0$.

If the discriminant of the second equation is negative, its roots are impossible (non-real).

So, if the roots of the first equation are possible and different, the roots of the second equation are impossible.

Part 2: Vice versa. If the roots of the second equation are impossible, then the roots of the first equation are possible and different.

The roots of the second equation are impossible means $D_2 < 0$.

$D_2 = -4k[(a + c)^2 + 4k] < 0$, where $k = b^2 - ac$.

Dividing by $-4$ and reversing the inequality sign, we get:

$k[(a + c)^2 + 4k] > 0$.

Let's consider the possibilities for $k = b^2 - ac$:

Case A: $k = 0$.

If $k = 0$, then $0 \times [(a + c)^2 + 4(0)] = 0$.

The inequality $k[(a + c)^2 + 4k] > 0$ becomes $0 > 0$, which is false.

So, $k$ cannot be 0 if $D_2 < 0$.

Case B: $k < 0$.

If $k < 0$, then for the product $k[(a + c)^2 + 4k]$ to be positive, the term $[(a + c)^2 + 4k]$ must be negative.

$(a + c)^2 + 4k < 0$

$(a + c)^2 < -4k$.

Since $k < 0$, $-4k > 0$. This inequality $(a + c)^2 < -4k$ is possible (e.g., if $a=1, c=1$, then $4 < -4k$, so $k < -1$).

If $k < 0$, the discriminant of the first equation $D_1 = 4k < 0$. This means the roots of the first equation are impossible (non-real).

However, the statement requires that if the roots of the second equation are impossible, then the roots of the first equation are possible and different ($k > 0$).

So, the case $k < 0$ is not consistent with the conclusion we want to reach ($k > 0$).

Case C: $k > 0$.

If $k > 0$, then $4k > 0$.

$(a + c)^2 \ge 0$.

So, $(a + c)^2 + 4k > 0 + 0 = 0$.

Thus, $k[(a + c)^2 + 4k]$ is the product of two positive numbers ($k > 0$ and $(a + c)^2 + 4k > 0$), which is positive.

$k[(a + c)^2 + 4k] > 0$.

This is consistent with the condition $D_2 < 0$.

So, if the roots of the second equation are impossible ($D_2 < 0$), then it must be that $k = b^2 - ac > 0$.

If $b^2 - ac > 0$, the discriminant of the first equation $D_1 = 4(b^2 - ac) > 0$.

This means the roots of the first equation are possible and different.

Combining both parts, we have shown that the roots of $ax^2 + 2bx + c = 0$ are possible and different if and only if the roots of $(a + c)(ax^2 + 2bx + c) = 2(ac - b^2) (x^2 + 1)$ are impossible. This proves the given statement.