I got stuck at finding $\beta$. Why did we use the product of roots?

Great question! You could have solved for $\beta$ by using the quadratic formula, but that would involve messy square roots. Using the product of roots (Vieta's formulas) is the elegant 'JEE way.' Since we already knew $\alpha^2$ and the product $\alpha\beta = -10$, squaring the product gave us $\alpha^2\beta^2 = 100$. It turns a complex algebraic problem into a simple arithmetic one. Always look for the path of least resistance!

JEE Main 2020 - 9 Jan (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: Let $a, b \in R, a \neq = 0$ be such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation, $x^{2} - 2 b x - 10 = 0$ . If $β$ is the other root of this equation, then $α^{2} + β^{2}$ is equal to:

Q: Wait, why do we use $x = -\frac{B}{2A}$ instead of the discriminant $D = 0$?

You absolutely can use the discriminant! Setting $D = B^2 - 4AC = 0$ is a perfectly valid and rigorous method. However, in competitive exams like JEE, time is your currency. Using $x = -\frac{B}{2A}$ is a direct shortcut derived from the vertex of the parabola. It saves you the algebraic headache of expanding $B^2 - 4AC$ and gets you to the value of $\alpha$ in one step. Both paths lead to the same truth, but one is a scenic route and the other is a high-speed rail.

Q: Why are we allowed to substitute $\alpha$ into the second equation?

The problem explicitly states that $\alpha$ is a root of *both* equations. In the language of algebra, this means $\alpha$ is a 'common root.' If a value satisfies an equation, it makes the equation true (equal to zero). Therefore, substituting $\alpha$ into the second equation is not just allowed; it is the key that unlocks the relationship between the variables $a$ and $b$.

Select Answer:

Visualized Solution

Analyze Equation 1: $a x^{2} - 2 b x + 5 = 0$

Given Equation 1: $a x^{2} - 2 b x + 5 = 0$
It has a repeated root $α$ .
Graphically, the parabola touches the x-axis at exactly one point.

Express $α$ in terms of $a$ and $b$

For a quadratic $A x^{2} + B x + C = 0$ , a repeated root is $x = - \frac{B}{2 A}$ .
Comparing coefficients: $A = a, B = - 2 b$ .
$α = - \frac{- 2 b}{2 a} = \frac{b}{a}$

Substitute $α$ into Equation 1

Since $α$ is a root, it satisfies $a α^{2} - 2 b α + 5 = 0$ .
Substitute $α = \frac{b}{a}$ :
$a (\frac{b}{a})^{2} - 2 b (\frac{b}{a}) + 5 = 0$

Simplify to find $b^{2}$ in terms of $a$

$\frac{a b ^{2}}{a ^{2}} - \frac{2 b ^{2}}{a} + 5 = 0$
$\frac{b ^{2}}{a} - \frac{2 b ^{2}}{a} + 5 = 0$
$- \frac{b ^{2}}{a} + 5 = 0 ⟹ b^{2} = 5 a$

Analyze Equation 2: $x^{2} - 2 b x - 10 = 0$

Given Equation 2: $x^{2} - 2 b x - 10 = 0$
Roots are $α$ and $β$ .
The parabola intersects the x-axis at $α$ and $β$ .

Substitute $α = \frac{b}{a}$ into Equation 2

Since $α$ is a root, $α^{2} - 2 b α - 10 = 0$ .
Substitute $α = \frac{b}{a}$ :
$(\frac{b}{a})^{2} - 2 b (\frac{b}{a}) - 10 = 0$

Substitute $b^{2} = 5 a$ to solve for $a$

$\frac{b ^{2}}{a ^{2}} - \frac{2 b ^{2}}{a} - 10 = 0$
Substitute $b^{2} = 5 a$ :
$\frac{5 a}{a ^{2}} - \frac{2 ( 5 a )}{a} - 10 = 0$

Calculate the value of $a$

$\frac{5}{a} - 10 - 10 = 0$
$\frac{5}{a} = 20$
$a = \frac{5}{20} = \frac{1}{4}$

Calculate $b^{2}$ and $α^{2}$

$b^{2} = 5 a = 5 (\frac{1}{4}) = \frac{5}{4}$
$α^{2} = (\frac{b}{a})^{2} = \frac{b ^{2}}{a ^{2}}$
$α^{2} = \frac{\frac{5}{4}}{\frac{1}{16}} = 20$

Find $β^{2}$ using Product of Roots

For $x^{2} - 2 b x - 10 = 0$ , product of roots $α β = - 10$ .
Squaring both sides: $α^{2} β^{2} = 100$ .
Substitute $α^{2} = 20$ :
$20 β^{2} = 100 ⟹ β^{2} = 5$

Final Calculation: $α^{2} + β^{2}$

We need to find $α^{2} + β^{2}$ .
$α^{2} = 20$ and $β^{2} = 5$
$α^{2} + β^{2} = 20 + 5 = 25$

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The Sigma Insight: Relation Between Roots and Coefficients

Solution Diagram

Analyzing the First Parabola

The first equation is given by

a x^{2} - 2 b x + 5 = 0

. We are told this equation has a repeated root,

α

For a quadratic equation

A x^{2} + B x + C = 0

to have a repeated root, its discriminant must be zero, or equivalently, the vertex must lie on the x-axis at

x = - \frac{B}{2 A}

Applying this to our equation where

A = a

and

B = - 2 b

, we find:

α = - \frac{- 2 b}{2 a} = \frac{b}{a}

Establishing the Relationship

Since

α

is a root, it must satisfy the equation

a x^{2} - 2 b x + 5 = 0

. Substituting

α = \frac{b}{a}

into the equation:

a (\frac{b}{a})^{2} - 2 b (\frac{b}{a}) + 5 = 0

Simplifying the terms:

\frac{b ^{2}}{a} - \frac{2 b ^{2}}{a} + 5 = 0

- \frac{b ^{2}}{a} + 5 = 0 ⟹ b^{2} = 5 a

Solving for Coefficients

The second equation is

x^{2} - 2 b x - 10 = 0

, which has roots

α

and

β

. Since

α

is a common root, we substitute

α = \frac{b}{a}

into this equation:

(\frac{b}{a})^{2} - 2 b (\frac{b}{a}) - 10 = 0

\frac{b ^{2}}{a ^{2}} - \frac{2 b ^{2}}{a} - 10 = 0

Substituting

b^{2} = 5 a

into the expression:

\frac{5 a}{a ^{2}} - \frac{2 ( 5 a )}{a} - 10 = 0

\frac{5}{a} - 10 - 10 = 0 ⟹ \frac{5}{a} = 20

Thus, we find

a = \frac{1}{4}

and

b^{2} = 5 (\frac{1}{4}) = \frac{5}{4}

Final Calculation

First, we calculate

α^{2}

α^{2} = (\frac{b}{a})^{2} = \frac{b ^{2}}{a ^{2}} = \frac{5/4}{1/16} = 20

For the second equation

x^{2} - 2 b x - 10 = 0

, the product of the roots is

α β = - 10

. Squaring both sides gives:

α^{2} β^{2} = 100

Substituting

α^{2} = 20

20 β^{2} = 100 ⟹ β^{2} = 5

The final sum is:

α^{2} + β^{2} = 20 + 5 = 25

Common Traps & FAQs

Wait, why do we use

x = - \frac{B}{2 A}

instead of the discriminant

D = 0

You absolutely can use the discriminant! Setting

D = B^{2} - 4 A C = 0

is a perfectly valid and rigorous method. However, in competitive exams like JEE, time is your currency. Using

x = - \frac{B}{2 A}

is a direct shortcut derived from the vertex of the parabola. It saves you the algebraic headache of expanding

B^{2} - 4 A C

and gets you to the value of

α

in one step. Both paths lead to the same truth, but one is a scenic route and the other is a high-speed rail.

Why are we allowed to substitute

α

into the second equation?

The problem explicitly states that

α

is a root of *both* equations. In the language of algebra, this means

α

is a 'common root.' If a value satisfies an equation, it makes the equation true (equal to zero). Therefore, substituting

α

into the second equation is not just allowed; it is the key that unlocks the relationship between the variables

a

and

b

I got stuck at finding

β

. Why did we use the product of roots?

Great question! You could have solved for

β

by using the quadratic formula, but that would involve messy square roots. Using the product of roots (Vieta's formulas) is the elegant 'JEE way.' Since we already knew

α^{2}

and the product

α β = - 10

, squaring the product gave us

α^{2} β^{2} = 100

. It turns a complex algebraic problem into a simple arithmetic one. Always look for the path of least resistance!

Select Answer:

Visualized Solution

Analyze Equation 1: $a x^{2} - 2 b x + 5 = 0$

Express $α$ in terms of $a$ and $b$

Substitute $α$ into Equation 1

Simplify to find $b^{2}$ in terms of $a$

Analyze Equation 2: $x^{2} - 2 b x - 10 = 0$

Substitute $α = \frac{b}{a}$ into Equation 2

Substitute $b^{2} = 5 a$ to solve for $a$

Calculate the value of $a$

Calculate $b^{2}$ and $α^{2}$

Find $β^{2}$ using Product of Roots

Final Calculation: $α^{2} + β^{2}$

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the First Parabola

Establishing the Relationship

Solving for Coefficients

Final Calculation

Similar Questions

Let $a, b \in R, a \neq = 0$ , such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation $x^{2} - 2 b x - 10 = 0$ . If $β$ is the root of this equation, then $α^{2} + β^{2}$ is equal to:

Let $a, b \in R$ be such that the equation $a x^{2} - 2 b x + 15 = 0$ has a repeated root $α$ . If $α$ and $β$ are the roots of the equation $x^{2} - 2 b x + 21 = 0$ , then $α^{2} + β^{2}$ 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 $α, β$ are roots of the equation $x^{2} + 5 (2) x + 10 = 0, α > β$ and $P_{n} = α^{n} - β^{n}$ for each positive integer $n$ , then the value of $\frac{P _{17} P _{20} + 5 2 P _{17} P _{19}}{P _{18} P _{19} + 5 2 P _{18}^{2}}$ is equal to

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 :

If $α, β$ are the roots of the equation $x^{2} - (5 + 3^{l o g_{3} 5} - 5^{l o g_{5} 3}) x + 3 (3^{(l o g_{3} 5)^{1/3}} - 5^{(l o g_{5} 3)^{2/3}} - 1) = 0$ then the equation, whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ , is

The numbers of pairs $(a, b)$ of real numbers, such that whenever $α$ is a root of the equation $x^{2} + a x + b = 0$ , $α^{2} - 2$ is also a root of this equation, is :

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Comprehension Passage

$a_{12} =$

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

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

Select Answer:

Visualized Solution

Analyze Equation 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ax2−2bx+5=0

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

Simplify to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2 in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Analyze Equation 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−2bx−10=0

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=ab​ into Equation 2

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2=5a to solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β2 using Product of Roots

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the First Parabola

Establishing the Relationship

Solving for Coefficients

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b∈R be such that the equation ax2−2bx+15=0 has a repeated root α. If α and β are the roots of the equation x2−2bx+21=0, then α2+β2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The numbers of pairs (a,b) of real numbers, such that whenever α is a root of the equation x2+ax+b=0, α2−2 is also a root of this equation, 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 (x−a)(x−b)=c,c=0. Then the roots of the equation (x−α)(x−β)+c=0 are

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 and β be two roots of the equation x2−64x+256=0. Then the value of (β5α3​)1/8+(α5β3​)1/8 is

Analyze Equation 1: $a x^{2} - 2 b x + 5 = 0$

Substitute $α$ into Equation 1

Simplify to find $b^{2}$ in terms of $a$

Analyze Equation 2: $x^{2} - 2 b x - 10 = 0$

Substitute $α = \frac{b}{a}$ into Equation 2

Substitute $b^{2} = 5 a$ to solve for $a$

Calculate the value of $a$

Calculate $b^{2}$ and $α^{2}$

Find $β^{2}$ using Product of Roots

Final Calculation: $α^{2} + β^{2}$

Let $a, b \in R$ be such that the equation $a x^{2} - 2 b x + 15 = 0$ has a repeated root $α$ . If $α$ and $β$ are the roots of the equation $x^{2} - 2 b x + 21 = 0$ , then $α^{2} + β^{2}$ is equal to:

The numbers of pairs $(a, b)$ of real numbers, such that whenever $α$ is a root of the equation $x^{2} + a x + b = 0$ , $α^{2} - 2$ is also a root of this equation, is :

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

$a_{12} =$

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

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