To solve the quadratic equation

$(a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0,$

we use the quadratic formula:

$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A},$

where $A$, $B$, and $C$ are the coefficients of $x^2$, $x$, and the constant term respectively. Here:

$A = a^2 + b^2, \quad B = 2(ac + bd), \quad C = c^2 + d^2.$

Step 1: Substitute into the quadratic formula

The roots are given by:

$x = \frac{-2(ac + bd) \pm \sqrt{[2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2)}}{2(a^2 + b^2)}.$

Step 2: Simplify the discriminant

The discriminant is:

$\Delta = [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2).$

Expanding $[2(ac + bd)]^2$:

$[2(ac + bd)]^2 = 4(ac + bd)^2 = 4(a^2c^2 + b^2d^2 + 2abcd).$

Now expand $4(a^2 + b^2)(c^2 + d^2)$:

$4(a^2 + b^2)(c^2 + d^2) = 4(a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2).$

Thus, the discriminant becomes:

$\Delta = 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2).$

Simplify:

$\Delta = 4(a^2c^2 + b^2d^2 + 2abcd - a^2c^2 - b^2d^2 - a^2d^2 - b^2c^2),$

$\Delta = 4(2abcd - a^2d^2 - b^2c^2).$

Step 3: Write the roots

The roots are:

$x = \frac{-2(ac + bd) \pm \sqrt{4(2abcd - a^2d^2 - b^2c^2)}}{2(a^2 + b^2)}.$

Simplify further:

$x = \frac{-(ac + bd) \pm \sqrt{2abcd - a^2d^2 - b^2c^2}}{a^2 + b^2}.$

Final Answer:

The roots of the equation are:

$x = \frac{-(ac + bd) \pm \sqrt{2abcd - a^2d^2 - b^2c^2}}{a^2 + b^2}.$

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Relative Questions:

1. What happens to the roots if $a, b, c, d$ are all equal?
2. How does the discriminant determine whether the roots are real or complex?
3. Can this be solved geometrically when $a = c$ and $b = d$?
4. What if $a^2 + b^2 = 0$? Does the quadratic equation still hold?
5. How does the symmetry of the coefficients $a, b, c, d$ affect the solution?

Tip:

Always check the discriminant ($\Delta$) to determine if the roots are real or complex before simplifying further.