Answer:
[tex]\dfrac12 < p < 5[/tex]
Step-by-step explanation:
To solve this, we need to use the discriminant.
Discriminant
[tex]b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0[/tex]
[tex]\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}[/tex]
[tex]\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}[/tex]
[tex]\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}[/tex]
If the given equation has 2 real roots, then we need to use:
[tex]b^2-4ac > 0[/tex]
Given equation: [tex]8x^2-4x+1-p=0[/tex]
[tex]\implies a=8, \quad b=-4, \quad c=(1-p)[/tex]
Substituting these values into [tex]b^2-4ac > 0[/tex] and solve for p:
[tex]\implies (-4)^2-4(8)(1-p) > 0[/tex]
[tex]\implies 16-32(1-p) > 0[/tex]
[tex]\implies 16-32+32p > 0[/tex]
[tex]\implies -16+32p > 0[/tex]
[tex]\implies 32p > 16[/tex]
[tex]\implies p > \dfrac{16}{32}[/tex]
[tex]\implies p > \dfrac12[/tex]
For the roots to be less than 1, first find the value of p when the root is 1.
If the root is 1, then [tex](x-1)[/tex] will be a factor, so [tex]f(1)=0[/tex]
Substitute [tex]x=1[/tex] into the given equation and solve for p:
[tex]\implies 8(1)^2-4(1)+1-p=0[/tex]
[tex]\implies 5-p=0[/tex]
[tex]\implies p=5[/tex]
Therefore, the values of p for which the given equation has two different real roots less than 1 are:
[tex]\dfrac12 < p < 5[/tex]
