Math, asked by darthvader1234onkarb, 8 months ago

I
(a) Write 4 V5, 5 v3 and 3 V7 in ascending order. .​

Answers

Answered by Anonymous
0

Step-by-step explanation:

\green{\bold{\underline{ ✪ UPSC-ASPIRANT✪ }}}

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

Q:-solve and verify the equation

 \frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{ 3} )

\huge\tt\underline\blue{Answer }

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⟹</p><p> \frac{1}{3} x - 4 = x  - ( \frac{1}{2}  +  \frac{x}{3} )

⟹</p><p> \frac{x}{3}  - 4 = x - ( \frac{3 + 2x}{6} )

⟹</p><p> \frac{x - 12}{3}  = x - ( \frac{2x + 3}{6} )

⟹</p><p> \frac{x - 12}{3}  = x -  \frac{2x - 3}{6}

⟹</p><p> \frac{x - 12}{3}  =  \frac{6x - 2x - 3}{6}

⟹</p><p> \frac{x - 12}{3}  =  \frac{4x - 3}{6}

cancelling 6( R.H.S) By 3 From L.H.S

⟹ \frac{x - 12}{1}  =  \frac{4x  - 3}{2} </p><p>

⟹</p><p>2(x - 12) = 4x - 3

⟹</p><p>2x - 24 = 4x - 3

⟹</p><p> - 24 + 3 = 4x - 2x

⟹</p><p> - 21 = 2x

⟹</p><p>x =  -  \frac{21}{2}

CHECK:-

⟹ \frac{  - \frac{21}{2} }{3}  - 4 =   - \frac{21}{2}  - ( \frac{1}{2}  + ( - ) \frac{ \frac{21}{2} }{3} )</p><p>

⟹</p><p> -  \frac{21}{6}  - 4 =  -  \frac{21}{2}  - ( \frac{1}{2}  -  \frac{21}{6} )

⟹</p><p>  - \frac{7}{2}  - 4 =   - \frac{21}{2} - ( \frac{1}{2}   -  \frac{7}{2} )

⟹</p><p> \frac{ - 7 - 8}{2}  = -   \frac{21}{2}  - ( -  \frac{6}{2} )

⟹ -  \frac{15}{2}  =  -  \frac{21}{2} - ( - 3) </p><p>

⟹</p><p>  - \frac{15}{2}  =  -  \frac{21}{2}  + 3

⟹</p><p> -  \frac{15}{2}  =  \frac{ - 21 + 6}{2}  =  -  \frac{15}{2}

THEREFORE,L.H.S=R.H.S

VERIFIED✔️

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HOPE IT HELPS YOU..

_____________________

Thankyou:)

Answered by sumanrudra22843
0

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

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