Question

What can you say about the roots of the following equations?

(i) $x^2 + 2(3a + 5)x + 2(9a^2 + 25) = 0$

(ii) $(y - a)(y - b) + (y - b)(y - c) + (y - c)(y - a) = 0$

Answer 1

(i) The roots are real and equal if $a=5/3$, non-real otherwise. (ii) The roots are always real, equal if $a=b=c$, distinct otherwise.

Answer 2

The nature of the roots of a quadratic equation $Ax^2 + Bx + C = 0$ is determined by its discriminant, $\Delta = B^2 - 4AC$.

• If $\Delta > 0$, the roots are real and distinct.
• If $\Delta = 0$, the roots are real and equal.
• If $\Delta < 0$, the roots are non-real (complex conjugate).
• If $\Delta \ge 0$, the roots are real.

Let's analyze each equation:

(i) $x^2 + 2(3a + 5)x + 2(9a^2 + 25) = 0$

This is a quadratic equation in $x$ with $A = 1$, $B = 2(3a + 5)$, and $C = 2(9a^2 + 25)$. The discriminant is $\Delta_1 = B^2 - 4AC$.

$\Delta_1 = [2(3a + 5)]^2 - 4(1)[2(9a^2 + 25)]$

$\Delta_1 = 4(3a + 5)^2 - 8(9a^2 + 25)$

$\Delta_1 = 4(9a^2 + 30a + 25) - 72a^2 - 200$

$\Delta_1 = 36a^2 + 120a + 100 - 72a^2 - 200$

$\Delta_1 = -36a^2 + 120a - 100$

Factor out -4:

$\Delta_1 = -4(9a^2 - 30a + 25)$

Recognize the perfect square trinomial $9a^2 - 30a + 25 = (3a)^2 - 2(3a)(5) + 5^2 = (3a - 5)^2$.

So, $\Delta_1 = -4(3a - 5)^2$.

Since $(3a - 5)^2 \ge 0$ for any real value of $a$, the discriminant $\Delta_1 = -4(3a - 5)^2 \le 0$.

• If $\Delta_1 = 0$, the roots are real and equal. This occurs when $(3a - 5)^2 = 0$, which means $3a - 5 = 0$, or $a = 5/3$.
• If $\Delta_1 < 0$, the roots are non-real (complex conjugate). This occurs when $(3a - 5)^2 > 0$, which means $3a - 5 \ne 0$, or $a \ne 5/3$.

Conclusion for (i): The roots are real and equal if $a = 5/3$, and non-real (complex) if $a \ne 5/3$.

(ii) $(y - a)(y - b) + (y - b)(y - c) + (y - c)(y - a) = 0$

Expand the terms:

$(y^2 - by - ay + ab) + (y^2 - cy - by + bc) + (y^2 - ay - cy + ca) = 0$

Combine like terms:

$(y^2 + y^2 + y^2) + (-ay - by - by - cy - cy - ay) + (ab + bc + ca) = 0$

$3y^2 - 2(a + b + c)y + (ab + bc + ca) = 0$

This is a quadratic equation in $y$ with $A = 3$, $B = -2(a + b + c)$, and $C = ab + bc + ca$. The discriminant is $\Delta_2 = B^2 - 4AC$.

$\Delta_2 = [-2(a + b + c)]^2 - 4(3)(ab + bc + ca)$

$\Delta_2 = 4(a + b + c)^2 - 12(ab + bc + ca)$

$\Delta_2 = 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 12(ab + bc + ca)$

$\Delta_2 = 4a^2 + 4b^2 + 4c^2 + 8ab + 8bc + 8ca - 12ab - 12bc - 12ca$

$\Delta_2 = 4a^2 + 4b^2 + 4c^2 - 4ab - 4bc - 4ca$

Factor out 2:

$\Delta_2 = 2(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)$

Rearrange the terms inside the parenthesis:

$\Delta_2 = 2[(a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ca + a^2)]$

$\Delta_2 = 2[(a - b)^2 + (b - c)^2 + (c - a)^2]$.

Since $(a - b)^2 \ge 0$, $(b - c)^2 \ge 0$, and $(c - a)^2 \ge 0$ for any real values of $a, b, c$, their sum is non-negative: $(a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0$. Therefore, the discriminant $\Delta_2 = 2[(a - b)^2 + (b - c)^2 + (c - a)^2] \ge 0$. Since $\Delta_2 \ge 0$, the roots are always real.

• If $\Delta_2 = 0$, the roots are real and equal. This occurs when $(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$, which happens if and only if $a - b = 0$, $b - c = 0$, and $c - a = 0$, i.e., $a = b = c$.
• If $\Delta_2 > 0$, the roots are real and distinct. This occurs when $(a - b)^2 + (b - c)^2 + (c - a)^2 > 0$, which happens if $a, b, c$ are not all equal.

Conclusion for (ii): The roots are always real. They are real and equal if $a = b = c$, and real and distinct if $a, b, c$ are not all equal.