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

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The discussion revolves around evaluating the expression $b_1^2 + 5b_2^2$ given the constraints $a_1^2 + 5a_2^2 = 10$, $a_2b_1 - a_1b_2 = 5$, and $a_1b_1 + 5a_2b_2 = \sqrt{105}$. Participants share their solutions and insights, highlighting the usefulness of specific identities in solving the problem. The conversation emphasizes collaboration and appreciation for each other's contributions. Ultimately, the focus remains on deriving the value of $b_1^2 + 5b_2^2$ through the provided equations.
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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 $b_1^2+5b_2^2$.
 
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anemone said:
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.

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.
 
mente oscura said:
Hello.

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:

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 $|a|=\sqrt{10}$.

$\therefore b_1^2+5b_2^2=|b|^2=23$
 
my Solution
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 $b_1^2 + 5b_2^2 = 23$
 
kaliprasad said:
my Solution
we have
$(p^2+q^2)(l^2 + m^2) = (pl-qm)^2+ (pm + ql)^2$
...

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