A-Level Edexcel Maths Pure Laws of Indices & Surds: The equation $(p

The equation $(p - 1)x^2 + 4x + (p - 5) = 0$, where $p$ is a constant has no real roots - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Question 7

The equation $(p - 1)x^2 + 4x + (p - 5) = 0$, where $p$ is a constant has no real roots. (a) Show that $p$ satisfies $p^2 - 6p + 1 > 0$ (b) Hence find the ... show full transcript

Worked Solution & Example Answer:The equation $(p - 1)x^2 + 4x + (p - 5) = 0$, where $p$ is a constant has no real roots - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Step 1

Show that $p$ satisfies $p^2 - 6p + 1 > 0$

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Answer

To demonstrate that the equation has no real roots, we need to evaluate the condition under which the discriminant is less than zero. The given equation is a quadratic in terms of $x$ . The discriminant $riangle$ is determined using the formula:

$riangle = b^2 - 4ac$

For our equation,

Coefficient of $x^2$ (a) = $p - 1$
Coefficient of $x$ (b) = $4$
Constant term (c) = $p - 5$

Thus, the discriminant is:

$riangle = 4^2 - 4(p - 1)(p - 5)$

Calculating it further:

$riangle = 16 - 4[(p - 1)(p - 5)]$
$= 16 - 4(p^2 - 6p + 5)$
$= 16 - 4p^2 + 24p - 20$
$= -4p^2 + 24p - 4$
$= -4(p^2 - 6p + 1)$

To ensure that the equation has no real roots, we want:
$riangle < 0$
Therefore,
$-4(p^2 - 6p + 1) < 0$
This implies:
$p^2 - 6p + 1 > 0$
Thus, we have shown the required inequality.

Step 2

Hence find the set of possible values of $p$

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Answer

To find the set of possible values of $p$ , we analyze the quadratic inequality:
$p^2 - 6p + 1 > 0$

We can find the roots of the corresponding equation, $p^2 - 6p + 1 = 0$ , using the quadratic formula:

$p = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a}$

Here, $a = 1$ , $b = -6$ , and $c = 1$ :

$p = rac{6 ext{±} ext{√}(6^2 - 4 imes 1 imes 1)}{2 imes 1}$
$= rac{6 ext{\pm} ext{\sqrt}(36 - 4)}{2}$
$= rac{6 ext{\pm} ext{\sqrt}(32)}{2}$
$= rac{6 ext{\pm} 4 ext{\sqrt}2}{2}$
$= 3 ext{±} 2 ext{√}2$

The roots are $p_1 = 3 - 2 ext{√}2$ and $p_2 = 3 + 2 ext{√}2$ .

The quadratic opens upwards (as the coefficient of $p^2$ is positive), indicating that it will be positive outside the interval formed by the roots. Thus, the solution for the inequality $p^2 - 6p + 1 > 0$ is:

$p < 3 - 2 ext{√}2 ext{ or } p > 3 + 2 ext{√}2$

Therefore, the set of possible values of $p$ is
$p < 3 - 2 ext{√}2 ext{ or } p > 3 + 2 ext{√}2$ .

The equation $(p - 1)x^2 + 4x + (p - 5) = 0$, where $p$ is a constant has no real roots - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Worked Solution & Example Answer:The equation $(p - 1)x^2 + 4x + (p - 5) = 0$, where $p$ is a constant has no real roots - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 1

Show that $p$ satisfies $p^2 - 6p + 1 > 0$

Hence find the set of possible values of $p$

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