рд╣рд┐рдВрджреА

If x2 + ЁЭСе12= 7 and x тЙа 0; find the value of: 7x3 + 8x тИТ 7ЁЭСе3 тИТ8ЁЭСе - Mathematics

Advertisements
Advertisements

рдкреНрд░рд╢реНрди

If x2 + `x^(1/2)`= 7 and  x ≠ 0; find the value of: 
7x3 + 8x − `7/x^3 - 8/x`

рдпреЛрдЧ
Advertisements

рдЙрддреНрддрд░

7x3 + 8x − `7/x^3 - 8/x`

terms with common factors: `7(x^3 - 1/x^3) + 8 (c-1/x)`

Finding the value of `(x-1/x)`

`(x-1/x)^2 = x^2 + 1/x^2 -2`

`(x-1/x)^2 = 7-2=5`

Therefore, `x-1/x = +-sqrt5`

`(x-1/x)^3 = x^3-1/x^3-3(x-1/x)`

`x^3-1/x^3 = (x-1/x)^3 +3(x-1/x)`

`x-1/x = +-sqrt5: x^3-1/x^3 = (+-sqrt5)^3 + 3(+-sqrt5)`

`=x^3 -1/x^3=+-5sqrt5 +-3sqrt5.x^3-1/x^3 = +-8sqrt5`

The values found for `(x-1/x) and (x^3-1/x^3)` are substituted into the expression from step `1:7 (+-8sqrt5) +8 (+-sqrt5).`

`7x^3+8x-7/x^3-8/x is +-64sqrt5`

shaalaa.com
Expansion of Formula
  рдХреНрдпрд╛ рдЗрд╕ рдкреНрд░рд╢реНрди рдпрд╛ рдЙрддреНрддрд░ рдореЗрдВ рдХреЛрдИ рддреНрд░реБрдЯрд┐ рд╣реИ?
рдЕрдзреНрдпрд╛рдп 4: Expansions (Including Substitution) - Exercise 4 (D) [рдкреГрд╖реНрда ремрек]

APPEARS IN

рд╕реЗрд▓рд┐рдирд╛ Concise Mathematics [English] Class 9 ICSE
рдЕрдзреНрдпрд╛рдп 4 Expansions (Including Substitution)
Exercise 4 (D) | Q 8 | рдкреГрд╖реНрда ремрек
Share
Notifications

Englishрд╣рд┐рдВрджреАрдорд░рд╛рдареА


      Forgot password?
Use app×