Expert answers to "Given a quadratic equation, find values of a..."

High School
Algebra
Algebra I
Quadratic function and equation

Leroy Davidson

Answered question

2022-03-25

Given a quadratic equation, find values of a such that the equation will have four different solutions.

2 a {(x + 1)}^{2} - | x + 1 | + 1 = 0

Answer & Explanation

aznluck4u72x4

Beginner2022-03-26Added 16 answers

Step 1
Let $t = | x + 1 |$ . Then
$2 a {(x + 1)}^{2} - | x + 1 | + 1 = 0 \Rightarrow 2 a t^{2} - t + 1 = 0$
t naturally has two real solutions, so if $2 a t^{2} - t + 1 = 0$
has two distinct real solutions, then the original equation will have four distinct real solutions.
For $2 a t^{2} - t + 1 = 0$ , we have
$Δ = {(- 1)}^{2} - 4 (2 a) (1) > 0$
$1 - 8 a > 0$
$a < \frac{18}{}$
But, we also need to consider that if a is negative, then the vertical reflection will cancel out the effect of the absolute value.
Therefore, we want $0 < a < \frac{1}{8}$

Boehm98wy

Beginner2022-03-27Added 18 answers

Step 1

2 a u^{2} = | u | - 1 \Rightarrow 2 a {| u |}^{2} - | u | + 1 = 0

then

| u | = \frac{1 \pm \sqrt{1 - 8 a}}{2 a}

so we choose

0 < a < \frac{18}{}

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