Question

Q:-solve this and then find pq
${a}^{4} + {b}^{4} = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$
solve this easy Question​

Answer 1

Step-by-step explanation:

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve this and then find pq

${a}^{4} + {b}^{4} = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$

$\huge\tt\underline\blue{Answer }$

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$⟹ {a}^{4} + {b}^{4} = { ({a}^{2} )}^{2} + {( {b}^{2} )}^{2}$

$⟹ {( {a}^{2} + {b}^{2} ) }^{2} - 4 {a}^{2} {b}^{2} </p><p></p><p>[tex]⟹ { ({a}^{2} + {b}^{2} )}^{2} - ( {2ab)}^{2}$

$⟹ {a}^{4} + {b}^{4} = ( {a}^{2} + {b}^{2} + 2ab)( {a}^{2} + {b}^{2} - 2ab)$

$⟹( {a}^{2} + 2ab + {b}^{2} )( {a}^{2} - 2ab + {b}^{2} ) = ( {a}^{2} + pab + {b}^{2} )( {a}^{2} - qab + {b}^{2} )$

On comparing both sides :-

we get p=2 & q =2

$∴pq = 2 {x}^{2} = 4$

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HOPE IT HELPS YOU..

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Answer 2

Step-by-step explanation:

f(x) = kx³ – 8x² + 5

Roots are α – β , α & α +β

Sum of roots = – (-8)/k

Sum of roots = α – β + α + α +β = 3α

= 3α = 8/k

= k = 8/3α

or we can solve as below

f(x) = (x – (α – β)(x – α)(x – (α +β))

= (x – α)(x² – x(α+β + α – β) + (α² – β²))

= (x – α)(x² – 2xα + (α² – β²))

= x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

= x³ – 3αx² + x(3α² – β²) + αβ² – α³

= kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

kx³ – 8x² + 5

k(3α² – β²) = 0 => 3α² = β²

k(αβ² – α³) = 5

=k(3α³ – α³) = 5

= k2α³ = 5

3αk = 8 => k = 8/3α

(8/3α)2α³ = 5

=> α² = 15/16

=> α = √15 / 4