+0

# make 'p' the subject.

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377
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+603

make "p" the subject.

$$x={5 \over3}- \sqrt{2p-1}$$

I did this:

$${3 \over5}x=- \sqrt{2p-1}$$

$$({3\over5}x)^2 = -2p-1$$

$$({3\over5}x)^2 + 1 = -2p$$

$${9 \over25}x^2 + 1 = -2p$$

$${34 \over25}x^2 = -2p$$

$$-{1 \over2}*{34 \over25}x^2 = p$$

$$-{17 \over25}x^2 = p$$

Is this correct?...Thank you for your time...

Nov 12, 2018
edited by juriemagic  Nov 12, 2018

#1
+24093
+10

make "p" the subject.
$$\displaystyle x={5 \over3}- \sqrt{2p-1}$$

$$\begin{array}{|rcll|} \hline x &=& \dfrac{5}{3} - \sqrt{2p-1} \quad & | \quad -\dfrac{5}{3} \\ x-\dfrac{5}{3} &=& - \sqrt{2p-1} \quad & | \quad \times(-1) \\ -x+\dfrac{5}{3} &=& \sqrt{2p-1} \\ \dfrac{5}{3}-x &=& \sqrt{2p-1} \quad & | \quad \text{square both sides} \\ \left(\dfrac{5}{3}-x\right)^2 &=& 2p-1 \quad & | \quad +1 \\ \left(\dfrac{5}{3}-x\right)^2+1 &=& 2p \quad & | \quad :2 \\ \mathbf{\dfrac{ \left(\dfrac{5}{3}-x\right)^2+1 } {2}} & \mathbf{=}& \mathbf{p} \\ \hline \end{array}$$

Nov 12, 2018

#1
+24093
+10

make "p" the subject.
$$\displaystyle x={5 \over3}- \sqrt{2p-1}$$

$$\begin{array}{|rcll|} \hline x &=& \dfrac{5}{3} - \sqrt{2p-1} \quad & | \quad -\dfrac{5}{3} \\ x-\dfrac{5}{3} &=& - \sqrt{2p-1} \quad & | \quad \times(-1) \\ -x+\dfrac{5}{3} &=& \sqrt{2p-1} \\ \dfrac{5}{3}-x &=& \sqrt{2p-1} \quad & | \quad \text{square both sides} \\ \left(\dfrac{5}{3}-x\right)^2 &=& 2p-1 \quad & | \quad +1 \\ \left(\dfrac{5}{3}-x\right)^2+1 &=& 2p \quad & | \quad :2 \\ \mathbf{\dfrac{ \left(\dfrac{5}{3}-x\right)^2+1 } {2}} & \mathbf{=}& \mathbf{p} \\ \hline \end{array}$$

heureka Nov 12, 2018
#2
+603
+2

Heureka,

Thank you so much for the answer....how daft of me, to multiply the fraction instead of subtracting it...gosh....thank you kindly..

juriemagic  Nov 12, 2018