System of equations

Adding both the equation..

$a^{2} + b^{2} = 37$

Subtracting both the equations...

$a \times b = -6$

$a = \frac{-6}{b}$

Substitutiong b in the first equation,

$a ^{2} + \frac{36}{a^{2}} = 37$

Multiplying by $a^{2}$ ,

$a^{4} - 37a^{2} + 36 = 0$

$(a^{2} - 36)(a^{2} - 1) = 0$

$a = \boxed{6, -6, 1, -1}$

Multiplying the solutions, we get the answer as \boxed{36}

Hello, we get $\begin{aligned} a^2 + b^2 &= 37 \\ 2ab &= 12 \end{aligned}$ so $a + b = \pm 5$ .

$a$ then satisfies $a^2 - 5a - 6$ or $a^2 + 5a - 6$ and by Vieta we have $(-6)(-6) = \boxed{36}$

Solving the system of equations: Subtracting the first to the second equation gives 2ab = - 12 and ab = -6. This also implies that a^2 + b^2 = 37. Since ab = -6, then b = -6/a. a^2 + (36/a^2) = 37. Multiplying both sides by a^2 gives... a^4 + 36 = 37a^2 Solving for the roots gives a = 6, -6, 1, -1. Hence the product of the roots is 36.

Let ${a}^2+{ab}+{b}={31} {.......(1)}$

and ${a}^2+{ab}+{b}={43} {.......(2)}$

from ${.......(2)}$

${b}^2={43}+{ab}-{a}^2$

substituting this into ${.......(1)}$ we get

${a}^2+{ab}+{43}+{ab}-{a}^2={31}$

from which we get ${2ab}={12}$

${ab}={6}$

the possible values of a are ${3},{2},{6} and {1}$

and their product ${3}\times{2}\times{6}\times{1}$

gives the answer of $\boxed{36}$

$a^2+ab+b^2=31, a^2-ab+b^2=43$ . Adding and subtracting these two equation we get $2a^2+2b^2=74 <=> a^2+b^2=37$ .......................(1)

$2ab=-12$ ................(2)

Now adding and subtracting the equations we get

$(a+b)^2=25 <=> (a+b)=\pm 5$

$(a-b)^2=49 <=> (a-b)=\pm 7$

Solving these equations we get a= ${6,-6,1,-1}$

The product of all possible values of a is $\boxed{36}$

Its weird that they don't give condition  that a & b are integers

why cant the value of a be in decimal?

+6,-6,+1 and -1

subtract equations and you get 2ab=-12 = ab=-6. so a= -6, a=-1, a=1, a=6 work (note that if you check -2,3 and 2,-3 they do not work). so multiply the a solutions and you get 36! (note: that is not a factorial sign, it's only a normal exclamation point)

We have two term a^2 + ab + b^2 = 31 and a^2 - ab + b^2 = 43

adding all term we have 2a^2 + 2b^2 = 74

<=> 2(a^2 + b^2) = 74

<=> a^2 + b^2 = 37

• a^2 + ab + b^2 = 31 then ab = -6

• a^2 - ab + b^2 = 43 then - ab = 6

we assume a is positive, because factor positive of 6 is {1, 2, 3, 6} and this product is even

then the product of a = 1 x 2 x 3 x 6 = 36

$a^{2}+ab+b^{2}=31 ---------------------- eq(1)$

$a^{2}-ab+b^{2}=43 ----------------------- eq(2)$

$eq(1)-eq(2),$

$2ab=-12$

and $ab=-6$

$So, b= \frac{-6}{a}$

$eq(1)+eq(2),$

$2(a^{2}+b^{2})=74$

And, $a^{2}+b^{2}=37$

$a^{2}+b^{2}=37 ------------------- eq(3)$

$2ab=-12 -----------------------------eq(4)$

$eq(3)+eq(4),$

$a^{2}+2ab+b^{2}=25$

$(a+b)^{2}=25$

$(a+b)=+5 or (a+b)=-5$

Since $b= \frac{-6}{a}$ ,

$a- \frac{6}{a} =+5$ and $a- \frac{6}{a} = -5$

$a^{2} - 6 = 5a$ and $a^{2} -6 = -5a$

Solve the above equations and you'll get

$a= 6, -6, 1, and -1$

Hence, product of all possible value is 36