Let $a, b \in \mathbb{R}$ be such that the equation $ax^2 - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2bx + 21 = 0$, then $\alpha^2 + \beta^2$ is equal to: Options: (A) 37 (B) 58 (C) 68 (D) 92

### Analyzing the First Quadratic Equation We are given the quadratic equation $ax^2 - 2bx + 15 = 0$. We are told that this equation possesses a repeated root, $\alpha$. By Vieta's formulas, the sum of the roots is given by: $$\alpha + \alpha = \frac{-(-2b)}{a} \Rightarrow 2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a}$$ The product of the roots for this same equation is: $$\alpha \cdot \alpha = \frac{15}{a} \Rightarrow \alpha^2 = \frac{15}{a}$$ ### Establishing the Relationship We now have two expressions involving $\alpha$ and $a$. By squaring the first relationship, $\alpha = \frac{b}{a}$, we obtain: $$\alpha^2 = \frac{b^2}{a^2}$$ Equating the two expressions for $\alpha^2$, we have: $$\frac{b^2}{a^2} = \frac{15}{a}$$ Assuming $a \neq 0$, we simplify this to find $a = \frac{b^2}{15}$. Substituting this back into our expression for $\alpha$, we get: $$\alpha = \frac{b}{a} = \frac{b}{\frac{b^2}{15}} = \frac{15}{b}$$ ### Solving for the Second Equation We now turn our attention to the second equation, $x^2 - 2bx + 21 = 0$. Since $\alpha$ is a common root, we substitute $\alpha = \frac{15}{b}$ into this equation: $$\left( \frac{15}{b} \right)^2 - 2b \left( \frac{15}{b} \right) + 21 = 0$$ Expanding the terms, we obtain: $$\frac{225}{b^2} - 30 + 21 = 0 \Rightarrow \frac{225}{b^2} - 9 = 0$$ Solving for $b^2$, we find: $$b^2 = \frac{225}{9} = 25$$ ### Final Calculation The second equation $x^2 - 2bx + 21 = 0$ has roots $\alpha$ and $\beta$. From Vieta's formulas, the sum of the roots is $\alpha + \beta = 2b$ and the product is $\alpha\beta = 21$. We seek the value of $\alpha^2 + \beta^2$. Using the algebraic identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, we substitute our known values: $$\alpha^2 + \beta^2 = (2b)^2 - 2(21) = 4b^2 - 42$$ Substituting $b^2 = 25$ into the expression: $$\alpha^2 + \beta^2 = 4(25) - 42 = 100 - 42 = 58$$ The final value is **58**.

JEE Main 2022 (25 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: 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:

Q: Why can't we just use the discriminant $D=0$ for the first equation?

You absolutely can! Setting the discriminant $D = (-2b)^2 - 4(a)(15) = 0$ gives $4b^2 - 60a = 0$, or $b^2 = 15a$. This is a perfectly valid path that leads to the same result. Sometimes, multiple roads lead to the same beautiful destination in mathematics.

Q: Wait, why do we reject the negative root here?

Actually, we don't reject anything! The problem asks for $\alpha^2 + \beta^2$, and since we are squaring the roots, the sign of $\alpha$ or $\beta$ doesn't change the final result. The symmetry of the expression $\alpha^2 + \beta^2$ makes it robust against sign variations.

Select Answer:

Visualized Solution

Problem Setup

Equation 1: $a x^{2} - 2 b x + 15 = 0$ (Repeated root $α$ )
Equation 2: $x^{2} - 2 b x + 21 = 0$ (Roots $α$ and $β$ )
Goal: Find the value of $α^{2} + β^{2}$

Sum of Roots for Equation 1

For $a x^{2} - 2 b x + 15 = 0$ , roots are $α, α$ .
Sum of roots: $α + α = \frac{- ( - 2 b )}{a}$
$2 α = \frac{2 b}{a} ⟹ α = \frac{b}{a}$

Product of Roots for Equation 1

Product of roots: $α \cdot α = \frac{15}{a}$
$α^{2} = \frac{15}{a}$

Eliminating $a$

From $α = \frac{b}{a}$ , squaring gives $α^{2} = \frac{b ^{2}}{a ^{2}}$ .
Equating the two expressions for $α^{2}$ :
$\frac{b ^{2}}{a ^{2}} = \frac{15}{a}$

Expressing $α$ in terms of $b$

Solving $\frac{b ^{2}}{a ^{2}} = \frac{15}{a}$ gives $a = \frac{b ^{2}}{15}$ .
Substitute $a$ back into $α = \frac{b}{a}$ :
$α = \frac{b}{\frac{b ^{2}}{15}} = \frac{15}{b}$

Substituting $α$ into Equation 2

$α$ is a root of $x^{2} - 2 b x + 21 = 0$ .
Substitute $α = \frac{15}{b}$ into the equation:
$(\frac{15}{b})^{2} - 2 b (\frac{15}{b}) + 21 = 0$

Calculating $b^{2}$

Expanding the terms:
$\frac{225}{b ^{2}} - 30 + 21 = 0$
$\frac{225}{b ^{2}} - 9 = 0 ⟹ \frac{225}{b ^{2}} = 9$
$b^{2} = \frac{225}{9} = 25$

Roots of Equation 2

For $x^{2} - 2 b x + 21 = 0$ , roots are $α, β$ .
Sum of roots: $α + β = 2 b$
Product of roots: $α β = 21$

Expanding $α^{2} + β^{2}$

We need to find $α^{2} + β^{2}$ .
Using the algebraic identity:
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
Substitute the sum and product:
$α^{2} + β^{2} = (2 b)^{2} - 2 (21)$

The Final Answer

Simplify the expression:
$α^{2} + β^{2} = 4 b^{2} - 42$
Substitute $b^{2} = 25$ :
$= 4 (25) - 42 = 100 - 42 = 58$
Final Answer: $58$

00:00 / 00:00

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the First Quadratic Equation

We are given the quadratic equation

a x^{2} - 2 b x + 15 = 0

. We are told that this equation possesses a repeated root,

α

By Vieta's formulas, the sum of the roots is given by:

α + α = \frac{- ( - 2 b )}{a} \Rightarrow 2 α = \frac{2 b}{a} \Rightarrow α = \frac{b}{a}

The product of the roots for this same equation is:

α \cdot α = \frac{15}{a} \Rightarrow α^{2} = \frac{15}{a}

Establishing the Relationship

We now have two expressions involving

α

and

a

. By squaring the first relationship,

α = \frac{b}{a}

, we obtain:

α^{2} = \frac{b ^{2}}{a ^{2}}

Equating the two expressions for

α^{2}

, we have:

\frac{b ^{2}}{a ^{2}} = \frac{15}{a}

Assuming $a eq 0$, we simplify this to find

a = \frac{b ^{2}}{15}

. Substituting this back into our expression for

α

, we get:

α = \frac{b}{a} = \frac{b}{\frac{b ^{2}}{15}} = \frac{15}{b}

Solving for the Second Equation

We now turn our attention to the second equation,

x^{2} - 2 b x + 21 = 0

. Since

α

is a common root, we substitute

α = \frac{15}{b}

into this equation:

(\frac{15}{b})^{2} - 2 b (\frac{15}{b}) + 21 = 0

Expanding the terms, we obtain:

\frac{225}{b ^{2}} - 30 + 21 = 0 \Rightarrow \frac{225}{b ^{2}} - 9 = 0

Solving for

b^{2}

, we find:

b^{2} = \frac{225}{9} = 25

Final Calculation

The second equation

x^{2} - 2 b x + 21 = 0

has roots

α

and

β

. From Vieta's formulas, the sum of the roots is

α + β = 2 b

and the product is

α β = 21

We seek the value of

α^{2} + β^{2}

. Using the algebraic identity

α^{2} + β^{2} = (α + β)^{2} - 2 α β

, we substitute our known values:

α^{2} + β^{2} = (2 b)^{2} - 2 (21) = 4 b^{2} - 42

Substituting

b^{2} = 25

into the expression:

α^{2} + β^{2} = 4 (25) - 42 = 100 - 42 = 58

The final value is 58.

Common Traps & FAQs

Why can't we just use the discriminant

D = 0

for the first equation?

You absolutely can! Setting the discriminant

D = (- 2 b)^{2} - 4 (a) (15) = 0

gives

4 b^{2} - 60 a = 0

, or

b^{2} = 15 a

. This is a perfectly valid path that leads to the same result. Sometimes, multiple roads lead to the same beautiful destination in mathematics.

Wait, why do we reject the negative root here?

Actually, we don't reject anything! The problem asks for

α^{2} + β^{2}

, and since we are squaring the roots, the sign of

α

β

doesn't change the final result. The symmetry of the expression

α^{2} + β^{2}

makes it robust against sign variations.

Why is the repeated root

α

so important?

The repeated root is the 'anchor' of the problem. It forces a specific relationship between the coefficients

a

and

b

. Without this constraint, the system would be underdetermined, and we wouldn't be able to find a unique value for

α^{2} + β^{2}

Animated Solution for Mathematics - Quadratic Equations: 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:

Select Answer:

Visualized Solution

Problem Setup

Sum of Roots for Equation 1

Product of Roots for Equation 1

Eliminating $a$

Expressing $α$ in terms of $b$

Substituting $α$ into Equation 2

Calculating $b^{2}$

Roots of Equation 2

Expanding $α^{2} + β^{2}$

The Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the First Quadratic Equation

Establishing the Relationship

Solving for the Second Equation

Final Calculation

Similar Questions

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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:

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Comprehension Passage

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Select Answer:

Visualized Solution

Problem Setup

Sum of Roots for Equation 1

Product of Roots for Equation 1

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

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

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

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

Roots of Equation 2

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

The Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the First Quadratic Equation

Establishing the Relationship

Solving for the Second Equation

Final Calculation

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