Question

Answer 2nd question

Answer 1

your question is wrong it should - root 2

Answer 2

(After editing all the errors in the question)

Solution:

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Given:

$4x^4 + 14x^3 - 38x^2 - 28x + 60 = 0$

x = √2, -√2,.

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To Find:

Other values of x

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We know that,

It is of the form,

$ax^4 + bx^3 + cx^2 + dx + e$,.

So,

a= 4,

b= 14,

c= -38,

d= -28,

e= 60,


We know that ,

=> α + β +γ + δ = $\frac{-b}{a}$  (Sum of the zeroes)

=> α + β + √2 +(-√2) = $\frac{-14}{4}$

=> α + β = -3.5 ..(i)

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=> αβγδ = $\frac{e}{a}$,

=> αβ(√2)(-√2) = $\frac{60}{4}$

=> αβ(-2) = 15

=> αβ= -7.5,..(ii)

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We know that,

=> (a + b)² = a² + 2ab + b²

=> (α + β)² = α² + 2αβ + β²

=> (-3.5)² = α² + 2(-7.5) + β²

=> 12.25 = α² -15 + β²

 =>27.25 = α² + β²,...(iii)

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=> (α - β)² = α² -2αβ + β²

=> (α - β)² = 27.25 - 2(-7.5)

=> (α - β)² = 27.25 + 15

=>(α - β)² = 42.25

=> α - β = 6.5 ...(iv)

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(i) + (iv) => α + β + α -β = - 3.5 + 6.5

=> 2α = 3

=> α = $\frac{3}{2}$

=> α = 1.5
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(i) => α + β = -3.5

=> 1.5 + β = -3.5

=> β = -3.5 - 1.5

=> β = -5,..

∴ The other zeroes are 1.5 & -6.

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                                           Hope it Helps!!