Find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ given that $a_1,a_2,a_3,a_4$ are roots of a biquadratic equation.

Question

So the biquadratic equation is $x^4+(2-\sqrt3)x^2+2+\sqrt3=0$. Let $a_1,a_2,a_3,a_4$ be its roots. So we have to find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ .
My attempt:
So of we put $x^2=t$, and let the roots of the new quadratic equation be $a_1,a_2$. So we get that $a_1=-a_3;a_2=-a_4$. So the expression whose value we have to find will become $(1-a_1^2)(1-a_2^2)$. And now we multiply the expression to get, $1-(a_1^2+a_2^2)+(a_1a_2)^2=1-((a_1+a_2)^2-2a_1a_2)+(a_1a_2)^2$.
Now substituting the values of sum of roots and product of the same, I get the answer $5+10\sqrt3$. But the answer according to my question paper is $5$.
So please can someone point out the mistake and tell the answer?

Answer 1

Here is your mistake:

As the new quadratic is a new equation, to avoid confusion I will use $b_1,b_2$ for the roots

$$1-(a_1^2+a_2^2)+(a_1a_2)^2=1-(b_1+b_2)+b_1b_2=1+(2-\sqrt{3})+2+\sqrt{3} $$

You used $a_1,a_2$ to denote both the original roots and their squares!!!!

Answer 2

Hint:

Just notice that $$(x-a_1)(x-a_2)(x-a_3)(x-a_4)=x^4+(2-\sqrt{3})x^2+2+\sqrt{3}$$

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About your attempt:

According to G-man (thanks) we have

\begin{align} 1-(a_1^2+a_2^2)+(a_1a_2)^2&=1-\left(-\frac{2-\sqrt{3}}{1}\right)+(2+\sqrt{3})\\ &=1+(2-\sqrt{3})+(2+\sqrt{3})\\ &=5 \end{align} Where we have used the fact that $(a_1a_2)^2=a_1a_2a_3a_4=2+\sqrt{3}$ from the Vieta's formulas.

Answer 3

Let $x=1-y$

So, we have $(1-y)^4+(2-\sqrt3)(1-y)^2+2+\sqrt3=0\iff y^4+\cdots+1+(2-\sqrt3)+(2+\sqrt3)=0$

$\implies\prod_{r=1}^4(1-a_r)=(-1)^4\dfrac{1+(2-\sqrt3)+(2+\sqrt3)}1$

Answer 4

You are confusing which are the roots. You have that $$t^2+(2-\sqrt 3)t+(2+\sqrt 3)=0$$

The roots of this equation are $a_1^2$ and $a_2^2$ and you want to find $$(1-a_1^2)(1-a_2^2)=1-(a_1^2+a_2^2)+a_1^2a_2^2$$

Now reading off the equation for $t$ you get $a_1^2+a_2^2=\sqrt 3-2$ and $a_1^2a_2^2=2+\sqrt 3$ and the answer comes out as $5$ as required.

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You don't need to go back to the equation for $x$ - that is where you got tangled up.

And the moral is: write everything out carefully if you get it wrong first time. It is so easy to have blind spots (I've had several I remember with shame)

Answer 5

\begin{align} f(x) &= (x-a_1)(x-a_2)(x-a_3)(x-a_4) \\ &= x^4 +(2−√3) x^2 +2 + √3 \\ f(1)& = 1 + 2 - √3 + 2 - √3 = 5 \end{align}