Question

Find a for which equation
$(a - 1)x^{4} + 4 {x}^{2} + (a + 2) = 0$
has all roots real ?
​

Answer 1

The solution is 1<a≤2.

Please put in brainliest...

Don't look at my picture!!! Last part!

Solution : Incomplete factorization?

Let the roots be $x^2=\alpha$ or $x^2=\beta$.

Because of x², α and β is real IF THEY ARE 0 OR POSITIVE.

We are being given :

α + β ≥ 0 ... 1

α × β ≥ 0 ... 2

D(discriminant) ≥ 0 since the solutions are real

We have :

$\alpha+\beta =-\frac{a+2}{a-1}$ ... 1'

$\alpha \times \beta =\frac{4}{a-1}$ ... 2'

$D=-4(a^2+a-6)$ ... 3' *Long calculation was removed.*

Now we have :

$\frac{a+2}{a-1}\leq 0$ ... 1'' Because of 1 & 1'

$\frac{4}{a-1}\geq 0$ ... 2'' Because of 2 & 2'

$(a+3)(a-2)\leq 0$ ... 3'' Because of 3 & 3'

Solution :

To use fraction inequality,

AB≤0 ↔ A≥0, B≤0 or opposite ... A

AB≥0 ↔ A≥0, B≥0 ... B

1'' ∴$a\geq 2$, $a<1$ / $1<a\leq 2$ Because denominator ≠ 0 and A

2'' ∴$a\geq 1$ Because of B

3'' ∴$-3\leq a\leq 2$ Because of quadratic function

Final solution : Find the common solution. (exhausted)

Don't need to consider $a\geq 2$, $a<1$ because share no common with 2''

Consider $-1<a\leq 2$.

Now you are free to look at my picture.

The solution is 1<a≤2.