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I have done the following for the median
$2(x-1) = .50$
results in $x = 1.25$

For the quartiles, I have $q_{1}$--> $2(x-1)=.25$ which results in $q_{1} = 1.125$
for $q_{3}$--> $2(x-1)=.75$ which results in $q_{3}=1.375$ and then subtract them to get $.25$ Please correct me if I am wrong! thank you in advance

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The median is the number $m$ for which $$\int_{x=-\infty}^m f(x) \, dx = \frac{1}{2}.$$ So you need to solve $$\frac{1}{2} = \int_{x=1}^m 2(x-1) \, dx = \left[x^2 - 2x\right]_{x=1}^m = m^2 - 2m - (1-2) = (m-1)^2.$$

For the IQR, you need to solve for the first and third quartiles in a similar fashion: $q_1$ satisfies $$\frac{1}{4} = \int_{x=1}^{q_1} f(x) \, dx,$$ and $q_3$ satisfies $$\frac{3}{4} = \int_{x=1}^{q_3} f(x) \, dx.$$ Then the IQR is simply $q_3 - q_1$.

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An approach using a beta distribution. Let $Y \sim \mathsf{Beta}(2,1).$ Then $X = Y+1.$ Results are the same as in @heropup's Answer (+1).

The quartiles of $Y$ are as follows (using R) are $1/2$ for the lower quartile, $\sqrt{2}/2$ for the median, and $\sqrt{3}/2$ for the upper quartile. The IQR is $(\sqrt{3} - 1)/2.$ Also, the lower quartile for $X$ is $1.5$ and the IQR for $X$ is the same as for $Y.$ [In R, qbeta is the quantile function (inverse CDF) of a beta distribution.]

qbeta(c(1,2,3)/4, 2, 1)
[1] 0.5000000 0.7071068 0.8660254
sqrt(1:3)/2
[1] 0.5000000 0.7071068 0.8660254

Simulation: By simulation of a million realizations of $X$ in R, we can get good approximations to the quartiles, accurate to about three significant digits--and a figure for illustration.

set.seed(223)
x = 1 + rbeta(10^6, 2, 1)
median(x);  IQR(x)
[1] 1.706684    # aprx 1.7071
[1] 0.3666178   # aprx 0.8660
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.001   1.499   1.707   1.666   1.866   2.000 

hdr = "Simulated values of 1+BETA(2,1) with Quartiles"
hist(x, prob=T, br=30, col="skyblue2", main=hdr)
 curve(dbeta(x-1, 2, 1), add=T, lwd=3, col="orange")
 abline(v=quantile(x, c(1:3)/4), col="darkgreen", lwd=2)

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