Question

Find the range of p in equation p(x+1) (x-3)=x-4p-2 has no real roots. ​

Answer 1

Answer:

$\displaystyle p > \frac{1}{4}$.

Step-by-step explanation:

Expand the left-hand side of this equation:

$p\, (x + 1)\, (x - 3) = p\, \left(x^2 - 2\, x - 3\right) = p\, x^2 - 2\, p\, x - 3\, p$.

Collect the terms, so that this quadratic equation is in the form $a\, x^2 + b\, x+ c = 0$:

$p\, x^2 - 2\, p\, x - 3\, p = x - 4\, p - 2$.

$p\, x^2 - (2\, p + 1)\, x + (p + 2) = 0$.

In this equation:

• $a = p$.
• $b = -(2\, p + 1)$.
• $c = p + 2$.

Calculate the quadratic discriminant of this quadratic equation:

$\begin{aligned}b^2 - 4\, a\, c &= (-(2\, p + 1))^2 - 4\, p\, (p + 2) \\ &= 4\, p^2 + 4\, p + 1 - 4\, p^2- 8\, p = -4\, p + 1\end{aligned}$.

A quadratic equation has no real root if its quadratic discriminant is less then zero. As a result, this quadratic equation will have no real root when $-4\, p + 1 < 0$. Solve for the range of $p$:

$\displaystyle p > \frac{1}{4}$.