Q.If $m$ is a number such that $1\le m\le 5$ (i.e. $m$ is equally likely to be any one of $1,2,3,4,5$), then the probability that the quadratic equation $2x^{2}+2mx+m+1=0$ has real roots is

Step 1. For $2x^2+2mx+(m+1)=0$ ($a=2,\ b=2m,\ c=m+1$), the discriminant is $D=b^2-4ac=(2m)^2-4(2)(m+1)=4m^2-8m-8$.

Step 2. Real roots require $D\ge 0$: $4m^2-8m-8\ge 0 \;\Rightarrow\; m^2-2m-2\ge 0$.

Step 3. Solve $m^2-2m-2=0$ by the quadratic formula: $m=\dfrac{2\pm\sqrt{4+8}}{2}=1\pm\sqrt{3}$, so $1-\sqrt3\approx -0.73$ and $1+\sqrt3\approx 2.73$. So $m^2-2m-2\ge0$ when $m\le 1-\sqrt3$ or $m\ge 1+\sqrt3$. …

Discriminant test ($b^2-4ac\ge0$) for real roots of the $m$-parametrised quadratic, applied to each equally-likely value of $m$ in the give …

PRINT NOTE: the source page (Exercise 12.5, Q25, both the rendered page image and the PDF text layer) prints only 'm <= 5' with no lower bound - the '1 <= ' half of the domain is missing from the book itself (a genuine textbook typesetting gap, confirmed on the page, not a text-extraction artifact). Without a lower bound the problem has no well-defined finite sample space. Testing candidate domains shows only the positive-integer domain {1,2,3,4,5} (i.e. 1<=m<=5) reproduces a probability that exactly matches one of the four printed options (3/5); no other natural reading (e.g. a continuous …