Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 205334 by cortano12 last updated on 17/Mar/24

Commented by Ghisom last updated on 17/Mar/24

$x^{2} - \frac{5}{2} x^{3 / 2} + 2 x - \frac{5}{4} x^{1 / 2} + \frac{1}{4} = 0$ $(x - 2 x^{1 / 2} + \frac{1}{2}) (x - \frac{1}{2} x^{1 / 2} + \frac{1}{2}) = 0$ $x = \frac{3}{2} \pm \sqrt{2}$ $x = - \frac{3}{8} \pm \frac{\sqrt{7}}{8} i$
	Answered by Rasheed.Sindhi last updated on 17/Mar/24
	$10(√x) +(5/( (√x) ))=4x+(1/x)+8 10(√x) +(5/( (√x) ))=(2(√x) )^2 +((1/( (√x) )))^2 +8 5(2(√x) +(1/( (√x) )))=(2(√x) +(1/( (√x) )))^2 +4 y^2 −5y+4=0 ; y=2(√x) +(1/( (√x) )) (y−1)(y−4)=0 y=1 or y=4 2(√x) +(1/( (√x) ))=1 or 2(√x) +(1/( (√x) ))=4 2x−(√x) +1=0 or 2x−4(√x) +1=0 { ((2((√x) )^2 −(√x) +1=0⇒(√x) =((1±i(√7))/4))),((2((√x) )^2 −4(√x) +1=0⇒(√x) =((4±2(√2))/4)=((2±(√2))/2))) :} { ((x=((−3±i(√7) )/8))),((x=((6±4(√2))/4)=((3±2(√2) )/2))) :}$
	$10 \sqrt{x} + \frac{5}{\sqrt{x}} = 4 x + \frac{1}{x} + 8$ $10 \sqrt{x} + \frac{5}{\sqrt{x}} = {(2 \sqrt{x})}^{2} + {(\frac{1}{\sqrt{x}})}^{2} + 8$ $5 (2 \sqrt{x} + \frac{1}{\sqrt{x}}) = {(2 \sqrt{x} + \frac{1}{\sqrt{x}})}^{2} + 4$ $y^{2} - 5 y + 4 = 0; y = 2 \sqrt{x} + \frac{1}{\sqrt{x}}$ $(y - 1) (y - 4) = 0$ $y = 1 or y = 4$ $2 \sqrt{x} + \frac{1}{\sqrt{x}} = 1 or 2 \sqrt{x} + \frac{1}{\sqrt{x}} = 4$ $2 x - \sqrt{x} + 1 = 0 or 2 x - 4 \sqrt{x} + 1 = 0$ ${\begin{cases} 2 {(\sqrt{x})}^{2} - \sqrt{x} + 1 = 0 \Rightarrow \sqrt{x} = \frac{1 \pm i \sqrt{7}}{4} \\ 2 {(\sqrt{x})}^{2} - 4 \sqrt{x} + 1 = 0 \Rightarrow \sqrt{x} = \frac{4 \pm 2 \sqrt{2}}{4} = \frac{2 \pm \sqrt{2}}{2} \end{cases}$ ${\begin{cases} x = \frac{- 3 \pm i \sqrt{7}}{8} \\ x = \frac{6 \pm 4 \sqrt{2}}{4} = \frac{3 \pm 2 \sqrt{2}}{2} \end{cases}$
	Commented by Rasheed.Sindhi last updated on 17/Mar/24

	$C o r r e c t e d s i r$
	Commented by cortano12 last updated on 17/Mar/24

	$10 \sqrt{x} + \frac{5}{\sqrt{x}} = 4 x + \frac{1}{x} + 8$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum