Question

IIT Jee Mains Question(Previous year)
Topic - Quadratic Equations.

Q - If
$\alpha \ne \beta$
and
${ \alpha }^{2} = 5 \alpha - 3$
and
${ \beta }^{2} = 5 \beta - 3$

then the equation havingα/β abd β/α as its roots is ...


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Answer 1

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Given :-

$\sf { \alpha }^{2} = 5 \alpha - 3 \\ \\ \bf \implies { \alpha }^{2} - 5 \alpha + 3 = 0 \: \: \: \: . \: . \: . \: \{i \} \\ \\ \sf { \beta }^{2} = 5 \beta - 3 \\ \\ \implies \bf { \beta }^{2} - 5 \beta + 3 = 0 \: \: \: \: . \: . \: . \: \{ii \}$

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To Find :-

Equation whose roots are α/β and β/α.

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Main Concept :-

If sum and product of zeros of any quadratic polynomial are s and p respectively,

Then,

The quadratic polynomial is given by :-

$\bf  {x}^{2}  - s \: x + p$

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Solution :-

Eqs (i) & (ii) proves that α and β are roots of

x² -5x + 3 = 0

Hence,

Sum of roots = α + β = 5

and

Product of roots = αβ = 3

Now,

$\sf \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ \sf = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ \sf = \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ \sf = \frac{25 - 6}{3} \\ \\ \implies \underline{ \boxed{ \bf\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{19}{13} }}$

And,

$\boxed{ \boxed{ \bf\frac{ \alpha }{ \beta } . \frac{ \beta }{ \alpha } = 1}}$

So,

for the required quadratic equation whose roots are

α/β and β/α :-

Sum of roots = 19/3

and

Product of roots = 1.

Hence,

the required quadratic equation will be :-

$\sf{x}^{2} - \frac{19}{3} x + 1 = 0 \\ \\ \implies \large \colorbox{skyblue}{ \underline{ \boxed{ \bf3 {x}^{2} - 19x + 3 = 0 }}}$

$\large\red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required \:  \: }}\\\huge\red{\mathfrak{\text{ A}nswer.}}$

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Answer 2

Given :-

If α  ≠ β and α² = 5α - 3 and β² = 5β - 3

To Find :-

then the equation having α/β abd β/α as its roots is ...

Solution :-

Given

α² = 5α - 3

β² = 5β - 3

Subtracting both

α² - β² = 5α - 3 - (5β - 3)

α² - β² = 5α - 3 + 5β + 3

(α + β)(α - β) = 5(α - β) - (3 + 3)

α + β = 5 (1)

Now

Add both

α² + β² = 5α - 3 + (5β - 3)

α² + β² = 5α - 3 + 5β - 3

α² + β² = 5(α + β) - 6

α² + β² = 5(5) - 6

α² + β² = 25 - 6

α² + β² = 19 (2)

Takin 2 as common

(α + β)² = α² + 2αβ + β²

(α + β)² = α² + β² + 2αβ

(α + β)² = 19 + 2αβ

(5)² = 19 + 2αβ

25 = 19 + 2αβ

25 - 19 = 2αβ

6 = 2αβ

6/2 = αβ

3 = αβ

Finding sum of zeroes

Sum = α/β + β/α

α² + β²/αβ

19/3

Finding product of zeroes

(α/β × β/α)

αβ/βα

1

Standard form of polynomial = x² - (α + β)x + αβ

x² - (19/3)x + 1

x² - 19x/3 + 1

3x² - 19x + 1