- #1
Albert1
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$a,b\in N$
given:
$(3a+b)^2+6a -2b =1544$
find:
$a+b=?$
given:
$(3a+b)^2+6a -2b =1544$
find:
$a+b=?$
Albert said:$a,b\in N$
given:
$(3a+b)^2+6a -2b =1544$
find:
$a+b=?$
nice solution (Yes)kaliprasad said:$(3a+b)^2 + 2(3a-b) = 1544$
add 4b+1 to both sides to get
$(3a+b)^2 + 2(3a+b) + 1 = 1545 + 4b$
or $(3a + b + 1)^2 = 1545 + 4b$
as 1545 mod 4 = 1 solution may exist
so we need to take odd squares above 1545
$(3a + b + 1) = 41 => 1545 + 4b = 1681 => b= 34$
this gives a = 2 or a + b = 36
$(3a + b + 1) = 43 => 1545 + 4b = 1849 => b= 76$ too large
so a+b = 36 is the only solution
The equation for finding $a+b$ given $(3a+b)^2+6a-2b=1544$ is $3a^2+6ab+b^2+6a-2b=1544$.
To solve for $a+b$, we can start by grouping like terms. This will give us $3a^2+(6a+6a)+b^2+(-2b-b)=1544$. Then, we can combine the like terms to get $3a^2+12a+b^2-3b=1544$. Next, we can use the quadratic formula to solve for $a$, giving us two solutions. We can then substitute these solutions into the original equation to solve for $b$. Finally, we can add the two solutions for $a$ and $b$ to find the value of $a+b$.
Yes, this equation can be solved algebraically by using the steps outlined in the previous answer. However, it may be easier to use a graphing calculator or a computer program to quickly find the solutions.
Yes, there are restrictions on the values of $a$ and $b$ in this equation. Since we are working with a quadratic equation, the value of $a$ cannot equal zero in order to have a valid solution. Additionally, the value of $b$ must be a multiple of $3$ in order to satisfy the equation $(3a+b)^2$. Other than these restrictions, any real numbers can be used for $a$ and $b$.
This equation can be applied in real-life situations that involve finding the sum of two unknown quantities that are related by a quadratic equation. For example, this equation may be used in physics to find the total force acting on an object when given the magnitude and direction of two forces. It can also be used in finance to find the total profit when given the cost and revenue of a business. Overall, this equation can be applied in any situation that involves finding the sum of two unknown quantities related by a quadratic equation.