If $\alpha+i\beta$ and $\gamma+i\delta$ are the roots of $x^{2}-(3-2i)x-(2i-2)=0, i=\sqrt{-1},$ then $\alpha\gamma+\beta\delta$ is equal to : Options: (A) -2 (B) 6 (C) -6 (D) 2

### Analyzing the Setup To solve the quadratic equation $x^{2}-(3-2i)x-(2i-2)=0$, we first map it to the standard form $ax^2 + bx + c = 0$. By comparing terms, we identify the coefficients as: $a = 1$ $b = -(3-2i)$ $c = -(2i-2) = 2-2i$ Being meticulous with these signs is essential to avoid errors. We will now deploy the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ ### The Discriminant: The Heart of the Engine Next, we calculate the discriminant $D = b^2 - 4ac$. First, we compute $b^2$: $$b^2 = [-(3-2i)]^2 = (3-2i)^2 = 9 + 4i^2 - 12i$$ Since $i^2 = -1$, this simplifies to $9 - 4 - 12i = 5 - 12i$. Next, we calculate $4ac$: $$4ac = 4(1)(2-2i) = 8 - 8i$$ Subtracting these values, we find the discriminant: $$D = (5 - 12i) - (8 - 8i) = -3 - 4i$$ ### The Square Root Challenge To proceed, we must find $\sqrt{-3-4i}$. We assume $\sqrt{-3-4i} = x+iy$. Squaring both sides yields: $$(x+iy)^2 = x^2 - y^2 + 2ixy = -3-4i$$ By comparing the real and imaginary parts, we solve for $x$ and $y$. This process reveals that: $$\sqrt{D} = \pm(1-2i)$$ ### The Final Reveal With $\sqrt{D}$ determined, we substitute back into the quadratic formula: $$x = \frac{(3-2i) \pm (1-2i)}{2}$$ This splits into two distinct cases for the roots: **Case 1 (Positive sign):** $$z_1 = \frac{(3-2i) + (1-2i)}{2} = \frac{4-4i}{2} = 2-2i$$ Here, we identify $\alpha = 2$ and $\beta = -2$. **Case 2 (Negative sign):** $$z_2 = \frac{(3-2i) - (1-2i)}{2} = \frac{2+0i}{2} = 1+0i$$ Here, we identify $\gamma = 1$ and $\delta = 0$. Finally, we evaluate the requested expression $\alpha\gamma + \beta\delta$: $$\alpha\gamma + \beta\delta = (2)(1) + (-2)(0) = 2$$ The final result is **2**.

JEE Main 2025 (January)

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Animated Solution for Mathematics - Complex Numbers: If $α + i β$ and $γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = - 1,$ then $α γ + β δ$ is equal to :

Q: Wait, why can't we just use the sum and product of roots formulas?

You absolutely can! The sum of roots is $\alpha + i\beta + \gamma + i\delta = -b/a = 3-2i$, and the product is $(\alpha + i\beta)(\gamma + i\delta) = c/a = 2-2i$. This gives you a system of equations. However, solving for $\alpha, \beta, \gamma, \delta$ individually from these can sometimes be more algebraically intensive than just finding the roots directly. Both paths are valid, but the quadratic formula is often more direct for this specific problem.

Q: How do I find the square root of a complex number like $-3-4i$?

Great question! Let $\sqrt{-3-4i} = x+iy$. Then $(x+iy)^2 = x^2 - y^2 + 2ixy = -3-4i$. Equating parts: $x^2 - y^2 = -3$ and $2xy = -4$. You can solve this system by substituting $y = -2/x$ into the first equation, leading to a biquadratic equation in $x$. It is a classic technique that unlocks the door to complex roots!

Select Answer:

Visualized Solution

Analyze the Equation

Given: $x^{2} - (3 - 2 i) x - (2 i - 2) = 0$
Roots: $z_{1} = α + i β$ and $z_{2} = γ + i δ$

Identify Coefficients

Standard form: $a x^{2} + b x + c = 0$
$a = 1$
$b = - (3 - 2 i)$
$c = - (2 i - 2) = 2 - 2 i$
Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Calculate $b^{2}$

$b^{2} = [- (3 - 2 i)]^{2} = (3 - 2 i)^{2}$
Expanding: $9 + 4 i^{2} - 12 i$
Since $i^{2} = - 1$ : $9 - 4 - 12 i = 5 - 12 i$

Calculate $4 a c$

$4 a c = 4 (1) (2 - 2 i)$
$4 a c = 8 - 8 i$

Find Discriminant $D$

$D = b^{2} - 4 a c$
$D = (5 - 12 i) - (8 - 8 i)$
$D = 5 - 8 - 12 i + 8 i = - 3 - 4 i$

Square Root of $D$

Need $D = - 3 - 4 i$
Let $(x + i y)^{2} = - 3 - 4 i$
Solving gives $D = \pm (1 - 2 i)$

Substitute into Formula

$x = \frac{- b \pm D}{2 a}$
$x = \frac{( 3 - 2 i ) \pm ( 1 - 2 i )}{2}$

Calculate First Root $z_{1}$

Taking the positive sign:
$z_{1} = \frac{( 3 - 2 i ) + ( 1 - 2 i )}{2}$
$z_{1} = \frac{4 - 4 i}{2} = 2 - 2 i$
So, $α = 2, β = - 2$

Calculate Second Root $z_{2}$

Taking the negative sign:
$z_{2} = \frac{( 3 - 2 i ) - ( 1 - 2 i )}{2}$
$z_{2} = \frac{2 + 0 i}{2} = 1 + 0 i$
So, $γ = 1, δ = 0$

Final Calculation

We need: $α γ + β δ$
Substitute values: $(2) (1) + (- 2) (0)$
Result: $2 + 0 = 2$

Conclusion

Final Answer: $2$
The quadratic formula is universally applicable, even with complex coefficients.

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

To solve the quadratic equation

x^{2} - (3 - 2 i) x - (2 i - 2) = 0

, we first map it to the standard form

a x^{2} + b x + c = 0

By comparing terms, we identify the coefficients as:

a = 1

b = - (3 - 2 i)

c = - (2 i - 2) = 2 - 2 i

Being meticulous with these signs is essential to avoid errors. We will now deploy the quadratic formula:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The Discriminant

The Heart of the Engine

Next, we calculate the discriminant

D = b^{2} - 4 a c

First, we compute

b^{2}

b^{2} = [- (3 - 2 i)]^{2} = (3 - 2 i)^{2} = 9 + 4 i^{2} - 12 i

Since

i^{2} = - 1

, this simplifies to

9 - 4 - 12 i = 5 - 12 i

Next, we calculate

4 a c

4 a c = 4 (1) (2 - 2 i) = 8 - 8 i

Subtracting these values, we find the discriminant:

D = (5 - 12 i) - (8 - 8 i) = - 3 - 4 i

The Square Root Challenge

To proceed, we must find

- 3 - 4 i

. We assume

- 3 - 4 i = x + i y

Squaring both sides yields:

(x + i y)^{2} = x^{2} - y^{2} + 2 i x y = - 3 - 4 i

By comparing the real and imaginary parts, we solve for

x

and

y

. This process reveals that:

D = \pm (1 - 2 i)

The Final Reveal

With

D

determined, we substitute back into the quadratic formula:

x = \frac{( 3 - 2 i ) \pm ( 1 - 2 i )}{2}

This splits into two distinct cases for the roots:

Case 1 (Positive sign):

z_{1} = \frac{( 3 - 2 i ) + ( 1 - 2 i )}{2} = \frac{4 - 4 i}{2} = 2 - 2 i

Here, we identify

α = 2

and

β = - 2

Case 2 (Negative sign):

z_{2} = \frac{( 3 - 2 i ) - ( 1 - 2 i )}{2} = \frac{2 + 0 i}{2} = 1 + 0 i

Here, we identify

γ = 1

and

δ = 0

Finally, we evaluate the requested expression

α γ + β δ

α γ + β δ = (2) (1) + (- 2) (0) = 2

The final result is 2.

Common Traps & FAQs

Wait, why can't we just use the sum and product of roots formulas?

You absolutely can! The sum of roots is

α + i β + γ + i δ = - b / a = 3 - 2 i

, and the product is

(α + i β) (γ + i δ) = c / a = 2 - 2 i

. This gives you a system of equations. However, solving for

α, β, γ, δ

individually from these can sometimes be more algebraically intensive than just finding the roots directly. Both paths are valid, but the quadratic formula is often more direct for this specific problem.

How do I find the square root of a complex number like

- 3 - 4 i

Great question! Let

- 3 - 4 i = x + i y

. Then

(x + i y)^{2} = x^{2} - y^{2} + 2 i x y = - 3 - 4 i

. Equating parts:

x^{2} - y^{2} = - 3

and

2 x y = - 4

. You can solve this system by substituting

y = - 2/ x

into the first equation, leading to a biquadratic equation in

x

. It is a classic technique that unlocks the door to complex roots!

Is the quadratic formula really valid for complex coefficients?

Yes! The derivation of the quadratic formula via completing the square does not rely on the coefficients being real. It only requires that the field of coefficients is a field where division by 2 is defined (which is true for complex numbers). So, it is perfectly safe to use.

Animated Solution for Mathematics - Complex Numbers: If $α + i β$ and $γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = - 1,$ then $α γ + β δ$ is equal to :

Select Answer:

Visualized Solution

Analyze the Equation

Identify Coefficients

Calculate $b^{2}$

Calculate $4 a c$

Find Discriminant $D$

Square Root of $D$

Substitute into Formula

Calculate First Root $z_{1}$

Calculate Second Root $z_{2}$

Final Calculation

Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Discriminant

The Square Root Challenge

The Final Reveal

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α+iβ and γ+iδ are the roots of x2−(3−2i)x−(2i−2)=0,i=−1​, then αγ+βδ is equal to :

Select Answer:

Visualized Solution

Analyze the Equation

Identify Coefficients

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4ac

Find Discriminant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Square Root of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Substitute into Formula

Calculate First Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​

Calculate Second Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​

Final Calculation

Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Discriminant

The Square Root Challenge

The Final Reveal

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: If $α + i β$ and $γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = - 1,$ then $α γ + β δ$ is equal to :

Calculate $b^{2}$

Calculate $4 a c$

Find Discriminant $D$

Square Root of $D$

Calculate First Root $z_{1}$

Calculate Second Root $z_{2}$

If $α, β \in R$ are such that $1 - 2 i$ (here $i^{2} = - 1$ ) is a root of $z^{2} + α z + β = 0$ , then $(α - β)$ is equal to:

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Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to :

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ where $i = - 1$ , then $16 \cdot R e (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}}) \cdot I m (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}})$ is equal to

Let $α$ and $β$ be the roots of the equation $x^{2} + (2 i - 1) = 0$ . Then, the value of $∣ α^{8} + β^{8} ∣$ is equal to :

If $α$ and $β$ be the roots of the equation $x^{2} - 2 x + 2 = 0$ , then the least value of $n$ for which $(\frac{α}{β})^{n} = 1$ is :

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $α$ and $β$ be the sum and the product of all the non-zero solutions of the equation $(\overset{z}{ˉ})^{2} + ∣ z ∣ = 0, z \in C$ . Then $4 (α^{2} + β^{2})$ is equal to :

If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is