We want to prove that this equation has no real roots under the condition ad ≠ bc.
The discriminant of a quadratic equation ax² + bx + c = 0 is given by D = b² - 4ac. If the discriminant is negative, the quadratic equation has no real roots.
In the given quadratic equation, the coefficients are:
We can use the discriminant of the quadratic equation to show that a c + b d = 0. The discriminant of a quadratic equation of the form ax^2 + bx + c = 0 is b^2 - 4ac. If the roots of the equation (a² + b²) x² + 2 (b c – a d) x + (c² + d²) = 0 are real and equal, then the discriminant must be equal to zero. Therefore, we have:(2 (b c – a d))^2 - 4 (a² + b²) (c² + d²) = 0 Simplifying this expression gives us:4 (b^2 c^2 - 2abcd + a^2 d^2) - 4 (a^2 c^2 + b^2 d^2) = 0 Rearranging the terms, we get:4 (b^2 c^2 - a^2 c^2 + a^2 d^2 - b^2 d^2 - 2abcd) = 0 Factoring out (a^2 + b^2), we have:4 (a^2 + b^2) (c^2 + d^2 - 2abcd/(a^2 + b^2)) = 0 Since a^2 + b^2 is positive, we can divide both sides by 4 (a^2 + b^2) to get:c^2 + d^2 - 2abcd/(a^2 + b^2) = 0 Multiplying both sides by (a^2 + b^2), we get:a^2 c^2 + b^2 c^2 + a^2 d^2 + b^2 d^2 - 2abcd = 0 Rearranging the terms, we get:(a^2 c^2 + b^2 d^2) + (a^2 d^2 + b^2 c^2) - 2abcd = 0 Using the identity (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2, we can rewrite the left-hand side as:(a^2 + b^2)(c^2 + d^2) - 2abcd = (ac + bd)^2 + (ad - bc)^2 - 2abcd Simplifying the right-hand side gives us:(ac + bd)^2 + (ad - bc)^2 - 2abcd = 0 Expanding the squares and simplifying, we get:a^2 c^2 + 2abcd + b^2 d^2 + a^2 d^2 - 2abcd + b^2 c^2 = 0 Simplifying further gives us:a^2(c^2 + d^2) + b^2(c^2 + d^2) = 0 Since the roots of the quadratic equation are real and equal, we know that (a² + b²) is positive, so we can divide both sides by (a² + b²) to get:c^2 + d^2 = 0 This implies that c = d = 0. Therefore, a c + b d = a(0) + b(0) = 0.
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Let's consider the quadratic equation:
(a² + b²)x² + 2(ac + bd)x + (c² + d²) = 0
We want to prove that this equation has no real roots under the condition ad ≠ bc.
The discriminant of a quadratic equation ax² + bx + c = 0 is given by D = b² - 4ac. If the discriminant is negative, the quadratic equation has no real roots.
In the given quadratic equation, the coefficients are:
a = a² + b²
b = 2(ac + bd)
c = c² + d²
Now, let's calculate the discriminant:
D = b² - 4ac
= (2(ac + bd))² - 4(a² + b²)(c² + d²)
Simplify the expression:
D = 4(a²c² + 2abcd + b²d²) - 4(a²c² + a²d² + b²c² + b²d²)
Simplify further:
D = 4a²c² + 8abcd + 4b²d² - 4a²c² - 4a²d² - 4b²c² - 4b²d²
Combine like terms:
D = -4a²d² - 4b²c² + 8abcd
Now, we know that ad ≠ bc. This implies that at least one of the terms -4a²d² or -4b²c² is nonzero, because if both were zero, then ad would equal bc.
Therefore, the discriminant D is a sum of two nonzero terms, one of which is positive and the other negative, resulting in a negative discriminant:
D = -4a²d² - 4b²c² + 8abcd < 0
Since the discriminant is negative, the quadratic equation (a² + b²)x² + 2(ac + bd)x + (c² + d²) = 0 has no real roots.
Hence, the equation has no real roots when ad ≠ bc.
Answer:
We can use the discriminant of the quadratic equation to show that a c + b d = 0. The discriminant of a quadratic equation of the form ax^2 + bx + c = 0 is b^2 - 4ac. If the roots of the equation (a² + b²) x² + 2 (b c – a d) x + (c² + d²) = 0 are real and equal, then the discriminant must be equal to zero. Therefore, we have:(2 (b c – a d))^2 - 4 (a² + b²) (c² + d²) = 0 Simplifying this expression gives us:4 (b^2 c^2 - 2abcd + a^2 d^2) - 4 (a^2 c^2 + b^2 d^2) = 0 Rearranging the terms, we get:4 (b^2 c^2 - a^2 c^2 + a^2 d^2 - b^2 d^2 - 2abcd) = 0 Factoring out (a^2 + b^2), we have:4 (a^2 + b^2) (c^2 + d^2 - 2abcd/(a^2 + b^2)) = 0 Since a^2 + b^2 is positive, we can divide both sides by 4 (a^2 + b^2) to get:c^2 + d^2 - 2abcd/(a^2 + b^2) = 0 Multiplying both sides by (a^2 + b^2), we get:a^2 c^2 + b^2 c^2 + a^2 d^2 + b^2 d^2 - 2abcd = 0 Rearranging the terms, we get:(a^2 c^2 + b^2 d^2) + (a^2 d^2 + b^2 c^2) - 2abcd = 0 Using the identity (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2, we can rewrite the left-hand side as:(a^2 + b^2)(c^2 + d^2) - 2abcd = (ac + bd)^2 + (ad - bc)^2 - 2abcd Simplifying the right-hand side gives us:(ac + bd)^2 + (ad - bc)^2 - 2abcd = 0 Expanding the squares and simplifying, we get:a^2 c^2 + 2abcd + b^2 d^2 + a^2 d^2 - 2abcd + b^2 c^2 = 0 Simplifying further gives us:a^2(c^2 + d^2) + b^2(c^2 + d^2) = 0 Since the roots of the quadratic equation are real and equal, we know that (a² + b²) is positive, so we can divide both sides by (a² + b²) to get:c^2 + d^2 = 0 This implies that c = d = 0. Therefore, a c + b d = a(0) + b(0) = 0.
Step-by-step explanation: