Find the value to three places of decimals of each of the following. It is given that
√2= 1.414,√3 = 1.732,√5 =2.236 and √10 = 3.162.
(i)2/√3 (ii)3/√10 (iii) √5 + 1 / √2 (iv) √10 + √15 / √2
(v) 2+ /√3 / 3 (vi) √2 - 1 / √5
Answers
Given : √2 = 1.414,√3 = 1.732,√5 = 2.236 and √10 = 3.162.
(i) We have 2/√3
On Rationalising the denominator :
= (2 × √3) /(√3 ×√3)
= 2√3/3
= (2 × 1.732)/3
= 3.464/3
= 1.154
Hence the value to three places of decimals of 2/√3 is 1.154.
(ii) We have 3/√10
On Rationalising the denominator :
= (3 × √10)/(√10 × √10)
= ( 3√10)/10
= (3 × 3.162)/10
= 9.486/10
= 0.9486
Hence the value to three places of decimals of 3/√10 is 0.9486 .
(iii) We have √5 + 1 / √2
On Rationalising the denominator :
[(√5 + 1) × √2] / )√2 × √2)
= (√10 + √2)/2
= (3.162 + 1.414)/2
= 4.576/2
= 2.288
Hence the value to three places of decimals of √5 + 1 / √2 is 2.288 .
(iv) We have √10 + √15 / √2
On Rationalising the denominator :
= [(√10 + √15) × √ 2] / (√2 × √2)
= [(√20 + √30)]/2
= [√(5× 4) + √3 × 10]/2
= [ 2√5 + √3 × √10]/2
= [2 × 2.236 + 1.732 × 3.162]/2
=[ 4.472 + 5.476584]/2
= 9.948584/2
= 4.974
Hence the value to three places of decimals of √10 + √15 / √2 is 4.974 .
(v) We have (2 + √3) / 3
= (2 + 1.732) / 3
= 3.732/3
= 1.244
Hence the value to three places of decimals of (2 + √3) / 3 is 1.244 .
(vi) We have (√2 - 1) / √5
On Rationalising the denominator :
= (√2 - 1) × √5 / √5 × √5
= (√10 - √5)/5
= (3.162 - 2.236)/5
= 0.926/5
= 0.185
Hence the value to three places of decimals of (√2 - 1) / √5 is 0.185 .
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Find the values of each of the following correct to three places of decimals, it being given that √2 = 1.414,√3 = 1.732, √5=2.236, √6= 2.4495 and √10= 3.162.
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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