+1124 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...
+26393 make "p" the subject.
\(\displaystyle x={5 \over3}- \sqrt{2p-1}\)
\(\begin{array}{|rcll|} \hline x &=& \dfrac{5}{3} - \sqrt{2p-1} \quad & | \quad -\dfrac{5}{3} \\ x-\dfrac{5}{3} &=& - \sqrt{2p-1} \quad & | \quad \times(-1) \\ -x+\dfrac{5}{3} &=& \sqrt{2p-1} \\ \dfrac{5}{3}-x &=& \sqrt{2p-1} \quad & | \quad \text{square both sides} \\ \left(\dfrac{5}{3}-x\right)^2 &=& 2p-1 \quad & | \quad +1 \\ \left(\dfrac{5}{3}-x\right)^2+1 &=& 2p \quad & | \quad :2 \\ \mathbf{\dfrac{ \left(\dfrac{5}{3}-x\right)^2+1 } {2}} & \mathbf{=}& \mathbf{p} \\ \hline \end{array}\)
+26393 make "p" the subject.
\(\displaystyle x={5 \over3}- \sqrt{2p-1}\)
\(\begin{array}{|rcll|} \hline x &=& \dfrac{5}{3} - \sqrt{2p-1} \quad & | \quad -\dfrac{5}{3} \\ x-\dfrac{5}{3} &=& - \sqrt{2p-1} \quad & | \quad \times(-1) \\ -x+\dfrac{5}{3} &=& \sqrt{2p-1} \\ \dfrac{5}{3}-x &=& \sqrt{2p-1} \quad & | \quad \text{square both sides} \\ \left(\dfrac{5}{3}-x\right)^2 &=& 2p-1 \quad & | \quad +1 \\ \left(\dfrac{5}{3}-x\right)^2+1 &=& 2p \quad & | \quad :2 \\ \mathbf{\dfrac{ \left(\dfrac{5}{3}-x\right)^2+1 } {2}} & \mathbf{=}& \mathbf{p} \\ \hline \end{array}\)
+1124 Heureka,
Thank you so much for the answer....how daft of me, to multiply the fraction instead of subtracting it...gosh..
..thank you kindly..
