JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: 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 ___

Enter Numerical Value:

Visualized Solution

Understanding the Objective

Given: $a + b + c = 1$
Given: $ab + b c + c a = 2$
Given: $ab c = 3$
Target: $a^{4} + b^{4} + c^{4}$

Identity for $a^{2} + b^{2} + c^{2}$

We need to step up from power $1$ to power $2$ .
Identity: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$
Rearranging: $a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + c a)$

Calculating $a^{2} + b^{2} + c^{2}$

Substitute $a + b + c = 1$ and $ab + b c + c a = 2$
$a^{2} + b^{2} + c^{2} = (1)^{2} - 2 (2)$
$a^{2} + b^{2} + c^{2} = 1 - 4 = - 3$

Targeting $a^{4} + b^{4} + c^{4}$

To get power $4$ , we square the sum of squares.
Identity: $(a^{2} + b^{2} + c^{2})^{2} = a^{4} + b^{4} + c^{4} + 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$
We need the value of $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}$ to proceed.

Identity for $(ab)^{2} + (b c)^{2} + (c a)^{2}$

Let's find $(ab)^{2} + (b c)^{2} + (c a)^{2}$ .
Square the pairwise sum: $(ab + b c + c a)^{2} = (ab)^{2} + (b c)^{2} + (c a)^{2} + 2 (ab \cdot b c + b c \cdot c a + c a \cdot ab)$
Factor out $ab c$ : $(ab + b c + c a)^{2} = (ab)^{2} + (b c)^{2} + (c a)^{2} + 2 ab c (a + b + c)$

Calculating $(ab)^{2} + (b c)^{2} + (c a)^{2}$

Substitute knowns: $ab + b c + c a = 2$ , $ab c = 3$ , $a + b + c = 1$
$(2)^{2} = (ab)^{2} + (b c)^{2} + (c a)^{2} + 2 (3) (1)$
$4 = (ab)^{2} + (b c)^{2} + (c a)^{2} + 6$
$(ab)^{2} + (b c)^{2} + (c a)^{2} = 4 - 6 = - 2$

Final Substitution

Recall from earlier: $(a^{2} + b^{2} + c^{2})^{2} = a^{4} + b^{4} + c^{4} + 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$
Substitute $a^{2} + b^{2} + c^{2} = - 3$ and $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = - 2$
$(- 3)^{2} = a^{4} + b^{4} + c^{4} + 2 (- 2)$

The Final Answer

Simplify the equation: $9 = a^{4} + b^{4} + c^{4} - 4$
Rearrange to solve: $a^{4} + b^{4} + c^{4} = 9 + 4$
Final Result: $a^{4} + b^{4} + c^{4} = 13$

00:00 / 00:00

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

We are given the elementary symmetric polynomials for three variables

a

b

, and

c

a + b + c = 1

ab + b c + c a = 2

ab c = 3

Our objective is to calculate the sum of the fourth powers,

a^{4} + b^{4} + c^{4}

Climbing to the Second Power

We begin by using the fundamental identity for the square of a trinomial:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)

Rearranging this to solve for the sum of squares, we have:

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

Substituting the given values:

a^{2} + b^{2} + c^{2} = (1)^{2} - 2 (2) = 1 - 4 = - 3

Finding the Intermediate Term

To reach the fourth power, we must evaluate the expression

(a^{2} + b^{2} + c^{2})^{2}

(a^{2} + b^{2} + c^{2})^{2} = a^{4} + b^{4} + c^{4} + 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})

We now need the value of

a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}

. We obtain this by squaring the pairwise sum:

(ab + b c + c a)^{2} = (ab)^{2} + (b c)^{2} + (c a)^{2} + 2 (ab \cdot b c + b c \cdot c a + c a \cdot ab)

Factoring out

ab c

from the last term, we get:

(ab + b c + c a)^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 ab c (a + b + c)

Substituting our known values into this equation:

(2)^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 (3) (1)

4 = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 6

a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = - 2

Final Calculation

Now, we substitute our results back into the expansion of the square of the sum of squares:

(- 3)^{2} = a^{4} + b^{4} + c^{4} + 2 (- 2)

Simplifying the arithmetic:

9 = a^{4} + b^{4} + c^{4} - 4

Solving for the final sum:

a^{4} + b^{4} + c^{4} = 9 + 4 = 13

The final answer is 13.

Common Traps & FAQs

Wait, the sum of squares

a^{2} + b^{2} + c^{2}

came out to be

- 3

. How can the sum of squares be negative? Did I make a mistake?

You have not made a mistake! In the realm of real numbers, squares are always non-negative. However, this result tells us that at least one of our variables

a, b,

c

must be a complex number. The algebraic identities we use are universal and hold true regardless of whether the variables are real or complex, so proceed with confidence!

Why can't I just solve for

a, b,

and

c

individually using the given equations and then raise them to the fourth power?

While theoretically possible, solving for the roots of a cubic equation is often a trap in JEE problems. It is computationally tedious and prone to error. Using symmetric polynomials is the 'elegant' path—it is faster, less error-prone, and demonstrates a deeper mastery of algebraic structures.

Why did we square the pairwise sum

(ab + b c + c a)

? It feels like we are just doing random algebra.

It is not random! We are playing a game of 'matching powers.' We need

a^{4} + b^{4} + c^{4}

. We know that squaring the sum of squares gives us these terms, but it also generates cross terms like

a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}

. Squaring the pairwise sum is the only direct way to generate those specific cross terms using the information we already have.

Animated Solution for Mathematics - Quadratic Equations: 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 ___

Enter Numerical Value:

Visualized Solution

Understanding the Objective

Identity for $a^{2} + b^{2} + c^{2}$

Calculating $a^{2} + b^{2} + c^{2}$

Targeting $a^{4} + b^{4} + c^{4}$

Identity for $(ab)^{2} + (b c)^{2} + (c a)^{2}$

Calculating $(ab)^{2} + (b c)^{2} + (c a)^{2}$

Final Substitution

The Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Climbing to the Second Power

Finding the Intermediate Term

Final Calculation

Similar Questions

Let $p$ and $q$ be two real numbers such that $p + q = 3$ and $p^{4} + q^{4} = 369$ . Then $(\frac{1}{p} + \frac{1}{q})^{- 2}$ 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

Let $a, b, c$ be real numbers with $a \neq = 0$ and let $α, β$ be the roots of the equation $a x^{2} + b x + c = 0$ . Express the roots of $a^{3} x^{2} + ab c x + c^{3} = 0$ in terms of $α, β$ .

If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 (α^{3} + β^{3})^{2}$ is equal to:

Let $α, β$ be the roots of the equation $x^{2} - 4 λ x + 5 = 0$ and $α, γ$ be the roots of the equation $x^{2} - (32 + 23) x + 7 + 3 λ 3 = 0$ . If $β + γ = 32$ , then $(α + 2 β + γ)^{2}$ is equal to ______.

If $α$ and $β$ be two roots of the equation $x^{2} - 64 x + 256 = 0$ . Then the value of $(\frac{α ^{3}}{β ^{5}})^{1/8} + (\frac{β ^{3}}{α ^{5}})^{1/8}$ is

Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β)$ . Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to

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

If $α, β$ are the roots of $a x^{2} + b x + c = 0, (a \neq = 0)$ and $α + δ, β + δ$ are the roots of $A x^{2} + B x + C = 0, (A \neq = 0)$ for some constant $δ$ , then prove that $(b^{2} - 4 a c) / a^{2} = (B^{2} - 4 A C) / A^{2}$

If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the $n$ -th power of the other, then show that $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$

Animated Solution for Mathematics - Quadratic Equations: .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 ___

Enter Numerical Value:

Visualized Solution

Understanding the Objective

Identity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2+c2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2+c2

Targeting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a4+b4+c4

Identity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (ab)2+(bc)2+(ca)2

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

Final Substitution

The Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Climbing to the Second Power

Finding the Intermediate Term

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 real numbers such that p+q=3 and p4+q4=369. Then (p1​+q1​)−2 is equal to ______.

.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; } Let a,b,c be real numbers with a=0 and let α,β be the roots of the equation ax2+bx+c=0. Express the roots of a3x2+abcx+c3=0 in terms of α,β.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the squares of the reciprocals of the roots α and β of the equation 3x2+λx−1=0 is 15, then 6(α3+β3)2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α and β be two roots of the equation x2−64x+256=0. Then the value of (β5α3​)1/8+(α5β3​)1/8 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β be the roots of the equation x2−x+2=0 with Im(α)>Im(β). Then α6+α4+β4−5α2 is equal to

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α,β are the roots of ax2+bx+c=0,(a=0) and α+δ,β+δ are the roots of Ax2+Bx+C=0,(A=0) for some constant δ, then prove that (b2−4ac)/a2=(B2−4AC)/A2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one root of the quadratic equation ax2+bx+c=0 is equal to the n-th power of the other, then show that (acn)1/(n+1)+(anc)1/(n+1)+b=0

Animated Solution for Mathematics - Quadratic Equations: 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 ___

Identity for $a^{2} + b^{2} + c^{2}$

Calculating $a^{2} + b^{2} + c^{2}$

Targeting $a^{4} + b^{4} + c^{4}$

Identity for $(ab)^{2} + (b c)^{2} + (c a)^{2}$

Calculating $(ab)^{2} + (b c)^{2} + (c a)^{2}$

Let $p$ and $q$ be two real numbers such that $p + q = 3$ and $p^{4} + q^{4} = 369$ . Then $(\frac{1}{p} + \frac{1}{q})^{- 2}$ 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

Let $a, b, c$ be real numbers with $a \neq = 0$ and let $α, β$ be the roots of the equation $a x^{2} + b x + c = 0$ . Express the roots of $a^{3} x^{2} + ab c x + c^{3} = 0$ in terms of $α, β$ .

If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 (α^{3} + β^{3})^{2}$ is equal to:

If $α$ and $β$ be two roots of the equation $x^{2} - 64 x + 256 = 0$ . Then the value of $(\frac{α ^{3}}{β ^{5}})^{1/8} + (\frac{β ^{3}}{α ^{5}})^{1/8}$ is

Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β)$ . Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to

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

If $α, β$ are the roots of $a x^{2} + b x + c = 0, (a \neq = 0)$ and $α + δ, β + δ$ are the roots of $A x^{2} + B x + C = 0, (A \neq = 0)$ for some constant $δ$ , then prove that $(b^{2} - 4 a c) / a^{2} = (B^{2} - 4 A C) / A^{2}$

If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the $n$ -th power of the other, then show that $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$