JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solution of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m$ . Then the value of $m$ is

Q: Why do we use $2i$ and $-2i$ specifically?

Great question! We need to create the term $(x_i^2 + 4)$. Notice that $(x_i - 2i)(x_i + 2i) = x_i^2 - (2i)^2 = x_i^2 - (-4) = x_i^2 + 4$. By evaluating the polynomial at $2i$ and $-2i$, we naturally generate these factors. It is a beautiful application of the factor theorem in the complex plane.

Q: Does the leading coefficient of $4$ matter?

It is absolutely critical! The factor theorem states $P(x) = a_n(x - x_1)(x - x_2)(x - x_3)(x - x_4)$. If you forget the leading coefficient $a_n = 4$, your final product will be off by a factor of $4^2 = 16$. Always keep an eye on the leading coefficient when working with roots.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Observe the structure of the target expression

(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2})

. Notice that

4 + x_{i}^{2}

can be written as

x_{i}^{2} - (2 i)^{2}

, which is a difference of squares:

(x_{i} - 2 i) (x_{i} + 2 i)

. This suggests that the product is related to the values of the polynomial at

x = 2 i

and

x = - 2 i

Hint 2

Hint 2 (The Arsenal): Recall the factored form of a polynomial

P (x) = a_{n} (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})

. If you evaluate

P (2 i)

and

P (- 2 i)

, you will generate factors of the form

(2 i - x_{i})

and

(- 2 i - x_{i})

. How can you combine these to form

(x_{i}^{2} + 4)

Hint 3

Hint 3 (The Execution): Calculate

P (2 i)

by substituting

x = 2 i

into the polynomial

4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9

. Since the coefficients are real,

P (- 2 i)

is simply the complex conjugate of

P (2 i)

. The product

P (2 i) P (- 2 i)

will then be equal to

16 \prod_{i = 1}^{4} (x_{i}^{2} + 4)

. Set this equal to

16 \times \frac{125}{16} m

and solve for

m

Visualized Solution

Define the Polynomial $P (x)$

Let the given polynomial be $P (x) = 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9$ .
The roots of the equation $P (x) = 0$ are given as $x_{1}, x_{2}, x_{3}, x_{4}$ .

Factorize $P (x)$ using Roots

Since $x_{1}, x_{2}, x_{3}, x_{4}$ are the roots, we can express $P (x)$ in factored form.
$P (x) = 4 (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})$

Analyze the Target Expression

We need to evaluate the product $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2})$ .
Notice that $4 + x_{i}^{2} = x_{i}^{2} - (2 i)^{2} = (x_{i} - 2 i) (x_{i} + 2 i)$ .

The Strategy: Evaluate $P (2 i)$ and $P (- 2 i)$

Consider the values of the polynomial at $x = 2 i$ and $x = - 2 i$ .
$P (2 i) = 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4})$
$P (- 2 i) = 4 (- 2 i - x_{1}) (- 2 i - x_{2}) (- 2 i - x_{3}) (- 2 i - x_{4})$

Calculating $P (2 i)$

Substitute $x = 2 i$ into $P (x) = 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9$ .
$P (2 i) = 4 (2 i)^{4} + 8 (2 i)^{3} - 17 (2 i)^{2} - 12 (2 i) + 9$
$P (2 i) = 4 (16) + 8 (- 8 i) - 17 (- 4) - 24 i + 9$
$P (2 i) = 64 - 64 i + 68 - 24 i + 9 = 141 - 88 i$

Calculating $P (- 2 i)$

Since the coefficients of $P (x)$ are real, $P (- 2 i)$ is the complex conjugate of $P (2 i)$ .
$P (- 2 i) = \overline{P (2 i)} = 141 + 88 i$

Product of $P (2 i)$ and $P (- 2 i)$

Calculate the product $P (2 i) P (- 2 i)$ .
$P (2 i) P (- 2 i) = (141 - 88 i) (141 + 88 i) = 14 1^{2} + 8 8^{2}$
$14 1^{2} = 19881$ , $8 8^{2} = 7744$
$P (2 i) P (- 2 i) = 19881 + 7744 = 27625$

Relating the Product to $m$

From the factored form: $P (2 i) P (- 2 i) = 16 \prod_{i = 1}^{4} (2 i - x_{i}) (- 2 i - x_{i})$
This simplifies to: $16 \prod_{i = 1}^{4} (4 + x_{i}^{2})$
Substitute the given value: $27625 = 16 (\frac{125}{16} m)$

Solving for $m$

Simplify the equation: $27625 = 16 (\frac{125}{16} m)$
The $16$ cancels out: $125 m = 27625$
$m = \frac{27625}{125} = 221$

Final Answer

The value of $m$ is $221$ .
Key Takeaway: Evaluating a polynomial $P (x)$ at complex points $ai$ and $- ai$ allows us to find the product of terms $(a^{2} + x_{i}^{2})$ efficiently.

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The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The given quartic equation is

P (x) = 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0

. This polynomial has four roots:

x_{1}, x_{2}, x_{3}, x_{4}

We aim to evaluate the product:

i = 1 \prod 4 (4 + x_{i}^{2})

The Complex Bridge

Observe that the term

(4 + x_{i}^{2})

can be factored using complex numbers. Since

4 = - (2 i)^{2}

, we can write:

4 + x_{i}^{2} = x_{i}^{2} - (2 i)^{2} = (x_{i} - 2 i) (x_{i} + 2 i)

Given the factored form of the polynomial

P (x) = 4 (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})

, we can evaluate the polynomial at

x = 2 i

and

x = - 2 i

P (2 i) = 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4})

P (- 2 i) = 4 (- 2 i - x_{1}) (- 2 i - x_{2}) (- 2 i - x_{3}) (- 2 i - x_{4})

The Calculation

Substituting

x = 2 i

into

P (x) = 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9

P (2 i) = 4 (2 i)^{4} + 8 (2 i)^{3} - 17 (2 i)^{2} - 12 (2 i) + 9

Using the properties

i^{2} = - 1

i^{3} = - i

, and

i^{4} = 1

, we simplify:

P (2 i) = 4 (16) + 8 (- 8 i) - 17 (- 4) - 24 i + 9

P (2 i) = 64 - 64 i + 68 - 24 i + 9 = 141 - 88 i

Since the coefficients of

P (x)

are real,

P (- 2 i)

is the complex conjugate of

P (2 i)

P (- 2 i) = 141 + 88 i

Multiplying these results yields:

P (2 i) P (- 2 i) = (141 - 88 i) (141 + 88 i) = 14 1^{2} + 8 8^{2}

P (2 i) P (- 2 i) = 19881 + 7744 = 27625

The Final Calculation

We know that:

P (2 i) P (- 2 i) = 16 i = 1 \prod 4 (2 i - x_{i}) (- 2 i - x_{i}) = 16 i = 1 \prod 4 (x_{i}^{2} + 4)

Given that the product is equal to

\frac{125}{16} m

, we set up the equation:

16 \times (\frac{125}{16} m) = 27625

125 m = 27625

Dividing both sides by

125

, we find the final value: $m = 221$

Common Traps & FAQs

Wait, why can't we just find the roots

x_{1}, x_{2}, x_{3}, x_{4}

and plug them in?

That is a natural instinct, but in JEE Advanced, we must look for the 'path of least resistance.' Finding the roots of a quartic equation is often algebraically nightmarish and prone to error. This problem is specifically designed to test your ability to use the properties of polynomials and complex numbers to bypass the calculation of individual roots entirely.

Why do we use

2 i

and

- 2 i

specifically?

Great question! We need to create the term

(x_{i}^{2} + 4)

. Notice that

(x_{i} - 2 i) (x_{i} + 2 i) = x_{i}^{2} - (2 i)^{2} = x_{i}^{2} - (- 4) = x_{i}^{2} + 4

. By evaluating the polynomial at

2 i

and

- 2 i

, we naturally generate these factors. It is a beautiful application of the factor theorem in the complex plane.

Does the leading coefficient of

4

matter?

It is absolutely critical! The factor theorem states

P (x) = a_{n} (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})

. If you forget the leading coefficient

a_{n} = 4

, your final product will be off by a factor of

4^{2} = 16

. Always keep an eye on the leading coefficient when working with roots.

Animated Solution for Mathematics - Quadratic Equations: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solution of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m$ . Then the value of $m$ is

Enter Numerical Value:

Visualized Solution

Define the Polynomial $P (x)$

Factorize $P (x)$ using Roots

Analyze the Target Expression

The Strategy: Evaluate $P (2 i)$ and $P (- 2 i)$

Calculating $P (2 i)$

Calculating $P (- 2 i)$

Product of $P (2 i)$ and $P (- 2 i)$

Relating the Product to $m$

Solving for $m$

Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Complex Bridge

The Calculation

The Final Calculation

Similar Questions

Let $p$ and $q$ be two positive number such that $p + q = 2$ and $p^{4} + q^{4} = 272$ . Then $p$ and $q$ are roots of the equation :

Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 57 i, P_{4} = - 37 i, P_{5} = 117 i$ and $P_{6} = 457 i,$ then $∣ α^{4} + β^{4} ∣$ is equal to

Let $α, β, γ$ be the three roots of the equation $x^{3} + b x + c = 0$ . If $β γ = 1 = - α$ , then $b^{3} + 2 c^{3} - 3 α^{3} - 6 β^{3} - 8 γ^{3}$ is equal to

The harmonic mean of the roots of the equation $(5 + 2) x^{2} - (4 + 5) x + 8 + 25 = 0$ is

If $λ$ be the ratio of the roots of the quadratic equation in $x$ , $3 m^{2} x^{2} + m (m - 4) x + 2 = 0$ , then the least value of $m$ for which $λ + 1/ λ = 1$ , is :

If $a + b + c = 1, ab + b c + c a = 2$ and $ab c = 3$ , then the value of $a^{4} + b^{4} + c^{4}$ is equal to ___

Let $α, β; α > β$ , be the roots of the equation $x^{2} - 2 x - 3 = 0$ . Let $P_{n} = α^{n} - β^{n}, n \in N$ . Then $(113 - 102) P_{10} + (112 + 10) P_{11} - 11 P_{12}$ is equal to

Comprehension Passage

$a_{12} =$

If $a_{4} = 28$ , then $p + 2 q =$

If $α, β$ , where $α < β$ , are the roots of the equation $λ x^{2} - (λ + 3) x + 3 = 0$ such that $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$ , then the sum of all possible values of $λ$ is

Let $λ \neq = 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0$ . Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation :

Enter Numerical Value:

Visualized Solution

Define the Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)

Factorize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) using Roots

Analyze the Target Expression

The Strategy: Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2i) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−2i)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2i)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−2i)

Product of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2i) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−2i)

Relating the Product to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

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

Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Complex Bridge

The Calculation

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p and q be two positive number such that p+q=2 and p4+q4=272. Then p and q are roots of the equation :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β,γ be the three roots of the equation x3+bx+c=0. If βγ=1=−α, then b3+2c3−3α3−6β3−8γ3 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The harmonic mean of the roots of the equation (5+2​)x2−(4+5​)x+8+25​=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If λ be the ratio of the roots of the quadratic equation in x, 3m2x2+m(m−4)x+2=0, then the least value of m for which λ+1/λ=1, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a+b+c=1,ab+bc+ca=2 and abc=3, then the value of a4+b4+c4 is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β;α>β, be the roots of the equation x2−2​x−3​=0. Let Pn​=αn−βn,n∈N. Then (113​−102​)P10​+(112​+10)P11​−11P12​ is equal to

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a12​=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a4​=28, then p+2q=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α,β, where α<β, are the roots of the equation λx2−(λ+3)x+3=0 such that α1​−β1​=31​, then the sum of all possible values of λ is

Define the Polynomial $P (x)$

Factorize $P (x)$ using Roots

The Strategy: Evaluate $P (2 i)$ and $P (- 2 i)$

Calculating $P (2 i)$

Calculating $P (- 2 i)$

Product of $P (2 i)$ and $P (- 2 i)$

Relating the Product to $m$

Solving for $m$

Let $p$ and $q$ be two positive number such that $p + q = 2$ and $p^{4} + q^{4} = 272$ . Then $p$ and $q$ are roots of the equation :

Let $α, β, γ$ be the three roots of the equation $x^{3} + b x + c = 0$ . If $β γ = 1 = - α$ , then $b^{3} + 2 c^{3} - 3 α^{3} - 6 β^{3} - 8 γ^{3}$ is equal to

The harmonic mean of the roots of the equation $(5 + 2) x^{2} - (4 + 5) x + 8 + 25 = 0$ is

If $λ$ be the ratio of the roots of the quadratic equation in $x$ , $3 m^{2} x^{2} + m (m - 4) x + 2 = 0$ , then the least value of $m$ for which $λ + 1/ λ = 1$ , is :

If $a + b + c = 1, ab + b c + c a = 2$ and $ab c = 3$ , then the value of $a^{4} + b^{4} + c^{4}$ is equal to ___

Let $α, β; α > β$ , be the roots of the equation $x^{2} - 2 x - 3 = 0$ . Let $P_{n} = α^{n} - β^{n}, n \in N$ . Then $(113 - 102) P_{10} + (112 + 10) P_{11} - 11 P_{12}$ is equal to

$a_{12} =$

If $a_{4} = 28$ , then $p + 2 q =$

If $α, β$ , where $α < β$ , are the roots of the equation $λ x^{2} - (λ + 3) x + 3 = 0$ such that $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$ , then the sum of all possible values of $λ$ is