Question

BASED ON ROOTS & FORMATION OF QUADRATIC EQUATION
If a, ß are roots of the equation 2x2 – 35 x + 2 = 0, then the value of (20 – 35)^3 . (2B - 35) ^3 is equal to-
(A) 1
(B) 8
(C) 64
(D) None of these
​

Answer 1

Question : -

If α , ß are roots of the equation 2x² - 35x + 2 = 0 , then the value of ( 2α - 35 )³ . ( 2ß - 35 )³ is equal to

( A ) 1

( B ) 8

( C ) 64

( D ) None of these

Answer : -

Given : -

α , ß are roots of the equation 2x² - 35x + 2 = 0

Required to find : -

• Value of ( 2α - 35 )³ . ( 2ß - 35 )³

Solution : -

2x² - 35x + 2 = 0

Here,

α , ß are roots of the equation

So,

This implies ;

x = α , x = ß

Now,

2x² - 35x + 2 = 0

( here , x = α )

2 ( α )² - 35 ( α ) + 2 = 0

2α² - 35a + 2 = 0

2a² - 35a = - 2

Taking α ( alpha ) common on the LHS part

α ( 2α - 35 ) = - 2

( 2α - 35 ) = -2/α $\longrightarrow{\tt{\green{\bf{ Equation - 1 }}}}$

Consider this as equation - 1

Similarly,

Now,

2x² - 35x + 2 = 0

( here , x = ß )

2 ( ß )² - 35 ( ß ) + 2 = 0

2ß² - 35ß + 2 = 0

2ß² - 35ß = - 2

Taking ß ( beta ) as common on LHS part

ß ( 2ß - 35 ) = - 2

2ß - 35 = -2/ß $\longrightarrow{\tt{\green{\bf{ Equation - 2 }}}}$

Consider this as Equation 2

Now,

Let's find the value of ( 2α - 35 )³ . ( 2ß - 35 )³

=> ( 2α - 35 )³ . ( 2ß - 35 )³

=> Substitute the values from equation 1 & equation 2

=> ( -2/α )³ . ( -2/ß )³

=> ( -2³/α³ ) . ( -2³/ß³ )

[ since, ( a/b )² = a²/b² ]

=> - 8/α³ x - 8/ß³

=> 64/α³ß³

=> 64/(αß)³

Now,

Let's find the value of αß

As we know that ;

$\: \boxed{\sf{ \pink{ \alpha + \beta = \dfrac{ - ( coefficient \; of \; x) }{ \: \: \: \: \: Coefficient \; of \; x^2 } }}}$

Similarly,

$\boxed{\sf{ \orange{ \alpha \beta = \dfrac{ Constant \; term }{ Coefficient \; of \; x^2 } }}}$

This implies ;

αß = constant term / coefficient of x²

αß = 2/2

αß = 1

Hence,

• Value of αß = 1

Substitute this value in the above calculations we get ;

=> 64/ ( 1 )³

=> 64/1

=> 64

Therefore,

Value of ( 2α - 35 )³ . ( 2ß - 35 )³ = 64

Option - c is correct ✓

Answer 2

$\mathcal{\huge{\fbox{\red{Question\:?}}}}$

✨ If a, ß are roots of the equation 2x2 – 35 x + 2 = 0, then the value of (20 – 35)³ . (2B - 35)³ is equal to-

(A) 1

(B) 8

(C) 64

(D) None of these

$\mathcal{\huge{\fbox{\green{AnSwEr:-}}}}$

➡ Option.c) 64

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

Given : -

• α , ß are roots of the equation, 2x² - 35x + 2 = 0.

To find : -

Value of ( 2α - 35 )³ × ( 2ß - 35 )³

Solution : -

➡ 2x² - 35x + 2 = 0 where, α , ß are roots of the equations.

So consider, x be α or x be ß,

In equation, 2x² - 35x + 2 = 0

Taking, x as α ,

➡ 2 ( α )² - 35 ( α ) + 2 = 0

➡ 2α² - 35a + 2 = 0

➡ 2a² - 35a = - 2

➡ α ( 2α - 35 ) = - 2

➡ ( 2α - 35 ) = -2/α ..............(1)

Again in equation, 2x² - 35x + 2 = 0

Now, taking x as β ,

➡ 2 ( ß )² - 35 ( ß ) + 2 = 0

➡ 2ß² - 35ß + 2 = 0

➡ 2ß² - 35ß = - 2

➡ ß ( 2ß - 35 ) = - 2

➡ 2ß - 35 = -2/ß.............(2)

Now,

Let's find the value of ( 2α - 35 )³ × ( 2ß - 35 )³

⟿ ( 2α - 35 )³ × ( 2ß - 35 )³

From equation 1 & 2 ,

⟿ ( -2/α )³ . ( -2/β )³

⟿ ( -2³/α³ ) . ( -2³/β³ ) [ where, ( a/b )² = a²/b² ]

⟿ - 8/α³ x - 8/β³

⟿ 64/α³β³

➮ 64/(αβ)³ ...............(3)

Using this formulae ⬇,

$α + β = \dfrac{-coefficient\:of\:x}{coefficient\:of\:x²}$

&

$αβ=\dfrac{Constant\:term}{Coefficient\:of\:x²}$

Now, αß = constant term / coefficient of x²

➡ αß = 2/2

➡ αß = 1

Hence,

Value of αß = 1

Putting in equation, 3

64/(αβ)³

⟿ 64/ ( 1 )³

⟿ 64/1

➠ 64

➣ The value of (2α – 35)³ × (2B - 35)³ is equal to 64 .

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