Question

For which values of the parameter "p" the equation

"8x^2 - 4x + 1 - p = 0" (with unknown x) has two different real roots less than 1?

Can you please explain in detail how this problem is solved?

Answer 1

`\frac12 < p < 5`

Explanation:

To solve this, we need to use the discriminant.

Discriminant

`b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0`

`\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}`

`\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}`

`\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}`

If the given equation has 2 real roots, then we need to use:

`b^2-4ac > 0`

Given equation:
`8x^2-4x+1-p=0`

`\implies a=8, \quad b=-4, \quad c=(1-p)`

Substituting these values into
`b^2-4ac > 0` and solve for p:

`\implies (-4)^2-4(8)(1-p) > 0`

`\implies 16-32(1-p) > 0`

`\implies 16-32+32p > 0`

`\implies -16+32p > 0`

`\implies 32p > 16`

`\implies p > (16)/(32)`

`\implies p > \frac12`

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
`(x-1)` will be a factor, so
`f(1)=0`

Substitute
`x=1` into the given equation and solve for p:

`\implies 8(1)^2-4(1)+1-p=0`

`\implies 5-p=0`

`\implies p=5`

Therefore, the values of p for which the given equation has two different real roots less than 1 are:

`\frac12 < p < 5`