Evaluate $b_1^2+5b_2^2$ Given $a_1^2+5a_2^2=10$

[anemone]

Given $a_1^2+5a_2^2=10,\,a_2b_1-a_1b_2=5$ and $a_1b_1+5a_2b_2=\sqrt{105}$ for $a_1,\,a_2,\,b_1,\,b_2\in R$, evaluate $b_1^2+5b_2^2$.

 

[mente oscura]

Given $a_1^2+5a_2^2=10,\,a_2b_1-a_1b_2=5$ and $a_1b_1+5a_2b_2=\sqrt{105}$ for $a_1,\,a_2,\,b_1,\,b_2\in R$, evaluate $b_1^2+5b_2^2$.

Hello.

Spoiler

$$x=a_1, \ y=a_2, \ z=b_1, \ k=b_2, \ z^2+5k^2=S$$

$$x^2+5y^2=10$$. (p1)

$$yz-xk=5$$. (p2)

$$xz+5yk= \sqrt{105}$$. (p3)

$$(x^2+5y^2)(z^2+5k^2)=10S$$. (pS)

$$y^2z^2+x^2k^2-2xyzk=25$$. Square (p2). (p4)

$$5y^2z^2+5x^2k^2-10xyzk=125$$. (p4)*5. (p5)

$$x^2z^2+25y^2k^2+10xyzk=105$$. Square (p3). (p6)

$$5y^2z^2+5x^2k^2+x^2z^2+25y^2k^2=230$$. (p5)+(p6). (p7)

$$(x^2+5y^2)(z^2+5k^2)=x^2z^2+5x^2k^2+5y^2z^2+25y^2k^2=10S=230$$

$$S=z^2+5k^2=23$$

$$b_1^2+5b_2^2=23$$

Regards.

 

[anemone]

Hello.

Spoiler

$$x=a_1, \ y=a_2, \ z=b_1, \ k=b_2, \ z^2+5k^2=S$$

$$x^2+5y^2=10$$. (p1)

$$yz-xk=5$$. (p2)

$$xz+5yk= \sqrt{105}$$. (p3)

$$(x^2+5y^2)(z^2+5k^2)=10S$$. (pS)

$$y^2z^2+x^2k^2-2xyzk=25$$. Square (p2). (p4)

$$5y^2z^2+5x^2k^2-10xyzk=125$$. (p4)*5. (p5)

$$x^2z^2+25y^2k^2+10xyzk=105$$. Square (p3). (p6)

$$5y^2z^2+5x^2k^2+x^2z^2+25y^2k^2=230$$. (p5)+(p6). (p7)

$$(x^2+5y^2)(z^2+5k^2)=x^2z^2+5x^2k^2+5y^2z^2+25y^2k^2=10S=230$$

$$S=z^2+5k^2=23$$

$$b_1^2+5b_2^2=23$$

Regards.

Thanks, mente oscura for your solution and thanks for participating too!

I will share the other solution here:

Spoiler

Let's define

$a=a_1+i\sqrt{5}a_2$

$b=b_1-i\sqrt{5}b_2$

The first given equation where $a_1^2+5a_2^2=10$ suggests $|a|=\sqrt{10}$

Multiply $a$ by $b$ we have

$\begin{align*}ab&=(a_1+i\sqrt{5}a_2)(b_1-i\sqrt{5}b_2)\\&=a_1b_1+5a_2b_2+i\sqrt{5}(a_2b_1-a_1b_2)\\&=\sqrt{105}+i\sqrt{5}5\end{align*}$

since we are told $a_2b_1-a_1b_2=5$ and $a_1b_1+5a_2b_2=\sqrt{105}$

This yields $|ab|=\sqrt{105+125}=\sqrt{230}=\sqrt{10}\sqrt{23}$ and this implies $|b|=\sqrt{23}$ since $|a|=\sqrt{10}$.

$\therefore b_1^2+5b_2^2=|b|^2=23$

 

[kaliprasad]

my Solution

Spoiler

we have
$(p^2+q^2)(l^2 + m^2) = (pl-qm)^2+ (pm + ql)^2$
putting $p= a_1$, $q = \sqrt{5} a_2$ ,$l= b_1$, $m = \sqrt{5} b_2$
we get $(a_1^2 + 5a_2^2)(b_1^2+5b_2)^2 = (a_1b_1 + 5a_2b_2)^2+5(a_1b_2 - a_2b_1)^2$
putting values from given conditions we get
$10(b_1^2+5b_2^2) = 105 + 5 * 5^2$
or $10(b_1^2 + 5b_2^2) = 230$
or $b_1^2 + 5b_2^2 = 23$

 

[anemone]

my Solution

Spoiler

we have
$(p^2+q^2)(l^2 + m^2) = (pl-qm)^2+ (pm + ql)^2$
...

Thanks, Kali for your solution and I especially want to give credit to the above "identity" because I can tell that is a lifesaver if we use it sparingly and wisely, hehehe...