Let $p, q \in \mathbb{R}$. If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^2 + px + q = 0$, then : Options: (A) $q^2 + 4p + 14 = 0$ (B) $p^2 - 4q - 12 = 0$ (C) $q^2 - 4p - 16 = 0$ (D) $p^2 - 4q + 12 = 0$

### Analyzing the Setup We are given the quadratic equation $x^2 + px + q = 0$ with one root $\alpha = 2 - \sqrt{3}$. In the realm of polynomials with rational coefficients, irrational roots always appear in conjugate pairs. This is known as the **Conjugate Root Theorem**. Since $\alpha = 2 - \sqrt{3}$ is a root, its conjugate $\beta = 2 + \sqrt{3}$ must also be a root of the equation. ### Applying Vieta's Formulas Vieta's formulas provide the bridge between the roots and the coefficients of the polynomial. For the equation $x^2 + px + q = 0$, the sum and product of the roots are defined as: $$\alpha + \beta = -p$$ $$\alpha \beta = q$$ ### Calculating the Coefficients First, we find the value of $p$ using the sum of the roots: $$\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$ Since $\alpha + \beta = -p$, we have $4 = -p$, which implies **$p = -4$**. Next, we find the value of $q$ using the product of the roots: $$\alpha \beta = (2 - \sqrt{3})(2 + \sqrt{3})$$ Applying the difference of squares identity $(a-b)(a+b) = a^2 - b^2$: $$\alpha \beta = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$$ Thus, **$q = 1$**. ### Final Verification To conclude, we evaluate the expression $p^2 - 4q - 12$ using our derived values: $$(-4)^2 - 4(1) - 12$$ $$= 16 - 4 - 12$$ $$= 0$$ The final result of the expression is **$0$**.

Why can't we just plug $x = 2 - \sqrt{3}$ into the equation $x^2 + px + q = 0$ and solve for $p$ and $q$?

You absolutely can! If you substitute $x = 2 - \sqrt{3}$ into $x^2 + px + q = 0$, you get $(2 - \sqrt{3})^2 + p(2 - \sqrt{3}) + q = 0$. Expanding this gives $(4 + 3 - 4\sqrt{3}) + 2p - p\sqrt{3} + q = 0$, which simplifies to $(7 + 2p + q) - \sqrt{3}(4 + p) = 0$. Since $\sqrt{3}$ is irrational, for this to be zero, both the rational part and the coefficient of $\sqrt{3}$ must be zero. This leads to $4 + p = 0$ and $7 + 2p + q = 0$. It is a powerful alternative method!

JEE Main 2019 (09 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: Let $p, q \in R$ . If $2 - 3$ is a root of the quadratic equation, $x^{2} + p x + q = 0$ , then :

Q: Wait, why do we assume the conjugate root exists? The problem says $p, q \in \mathbb{R}$, not necessarily rational.

That is a brilliant observation! In a general sense, if $p$ and $q$ are just any real numbers, the conjugate root theorem doesn't strictly apply. However, in the context of JEE problems where options are algebraic expressions involving integers, it is implicitly understood that the coefficients are rational. Without this assumption, the problem would be underspecified. Always look for the 'hidden' rational constraint in such competitive problems!

Q: Is there a faster way to check the options?

Once you have $p = -4$ and $q = 1$, you can simply plug these values into the options. It is the most reliable way to ensure you haven't made a sign error. Never rush the final verification step—it is where the most marks are lost!

Select Answer:

Visualized Solution

Quadratic Equation

Given equation: $x^{2} + p x + q = 0$
Coefficients: $p, q \in R$

Given Root $α$

One root is given as: $α = 2 - 3$
This is an irrational root.

Conjugate Root Theorem

Assuming rational coefficients, irrational roots occur in conjugate pairs.
If $a - b$ is a root, then $a + b$ is also a root.

Second Root $β$

By the conjugate pair rule, the second root is:
$β = 2 + 3$

Sum of Roots

Sum of roots: $α + β = - \frac{coefficient of x}{coefficient of x ^{2}}$
$α + β = - p$

Calculating $p$

Substitute the roots:
$(2 - 3) + (2 + 3) = - p$
$4 = - p ⟹ p = - 4$

Product of Roots

Product of roots: $α β = \frac{constant term}{coefficient of x ^{2}}$
$α β = q$

Calculating $q$

Substitute the roots:
$(2 - 3) (2 + 3) = q$
$2^{2} - (3)^{2} = q$
$4 - 3 = 1 ⟹ q = 1$

Testing the Options

We have found: $p = - 4$ and $q = 1$
Let's test Option 2: $p^{2} - 4 q - 12 = 0$

Verification

Substitute $p = - 4$ and $q = 1$ into Option 2:
$(- 4)^{2} - 4 (1) - 12$
$16 - 4 - 12 = 0$
Option 2 is correct.

00:00 / 00:00

The Sigma Insight: Nature of Roots

Solution Diagram

Analyzing the Setup

We are given the quadratic equation

x^{2} + p x + q = 0

with one root

α = 2 - 3

In the realm of polynomials with rational coefficients, irrational roots always appear in conjugate pairs. This is known as the Conjugate Root Theorem.

Since

α = 2 - 3

is a root, its conjugate

β = 2 + 3

must also be a root of the equation.

Applying Vieta's Formulas

Vieta's formulas provide the bridge between the roots and the coefficients of the polynomial. For the equation

x^{2} + p x + q = 0

, the sum and product of the roots are defined as:

α + β = - p

α β = q

Calculating the Coefficients

First, we find the value of

p

using the sum of the roots:

α + β = (2 - 3) + (2 + 3) = 4

Since

α + β = - p

, we have

4 = - p

, which implies $p = - 4$ .

Next, we find the value of

q

using the product of the roots:

α β = (2 - 3) (2 + 3)

Applying the difference of squares identity

(a - b) (a + b) = a^{2} - b^{2}

α β = 2^{2} - (3)^{2} = 4 - 3 = 1

Thus, $q = 1$ .

Final Verification

To conclude, we evaluate the expression

p^{2} - 4 q - 12

using our derived values:

(- 4)^{2} - 4 (1) - 12

= 16 - 4 - 12

= 0

The final result of the expression is $0$ .

Common Traps & FAQs

Wait, why do we assume the conjugate root exists? The problem says

p, q \in R

, not necessarily rational.

That is a brilliant observation! In a general sense, if

p

and

q

are just any real numbers, the conjugate root theorem doesn't strictly apply. However, in the context of JEE problems where options are algebraic expressions involving integers, it is implicitly understood that the coefficients are rational. Without this assumption, the problem would be underspecified. Always look for the 'hidden' rational constraint in such competitive problems!

Why can't we just plug

x = 2 - 3

into the equation

x^{2} + p x + q = 0

and solve for

p

and

q

You absolutely can! If you substitute

x = 2 - 3

into

x^{2} + p x + q = 0

, you get

(2 - 3)^{2} + p (2 - 3) + q = 0

. Expanding this gives

(4 + 3 - 43) + 2 p - p 3 + q = 0

, which simplifies to

(7 + 2 p + q) - 3 (4 + p) = 0

. Since

3

is irrational, for this to be zero, both the rational part and the coefficient of

3

must be zero. This leads to

4 + p = 0

and

7 + 2 p + q = 0

. It is a powerful alternative method!

Is there a faster way to check the options?

Once you have

p = - 4

and

q = 1

, you can simply plug these values into the options. It is the most reliable way to ensure you haven't made a sign error. Never rush the final verification step—it is where the most marks are lost!

Animated Solution for Mathematics - Quadratic Equations: Let $p, q \in R$ . If $2 - 3$ is a root of the quadratic equation, $x^{2} + p x + q = 0$ , then :

Select Answer:

Visualized Solution

Quadratic Equation

Given Root $α$

Conjugate Root Theorem

Second Root $β$

Sum of Roots

Calculating $p$

Product of Roots

Calculating $q$

Testing the Options

Verification

The Sigma Insight: Nature of Roots

Analyzing the Setup

Applying Vieta's Formulas

Calculating the Coefficients

Final Verification

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p,q∈R. If 2−3​ is a root of the quadratic equation, x2+px+q=0, then :

Select Answer:

Visualized Solution

Quadratic Equation

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

Conjugate Root Theorem

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

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