Question

If $2a=3\sqrt{a}$, $(∵a≠0)$ is expressed in a standard form of a quadratic equation, then a + b + c is __.

Answer 1

-5

Answer 2

To express the given equation $2a = 3\sqrt{a}$ in the standard form of a quadratic equation, $Ax^2 + Bx + C = 0$, we follow these steps:

1. Eliminate the square root: Square both sides of the equation. $(2a)^2 = (3\sqrt{a})^2$ $4a^2 = 9a$

2. Rearrange into standard form: Move all terms to one side to set the equation equal to zero. $4a^2 - 9a = 0$

3. Identify coefficients: The standard form of a quadratic equation is typically written as $Ax^2 + Bx + C = 0$. In this case, the variable is 'a'. So, comparing $4a^2 - 9a + 0 = 0$ with $Aa^2 + Ba + C = 0$: $A = 4$ $B = -9$ $C = 0$

4. Calculate the sum of coefficients: The question asks for $a+b+c$. It is implied that $a, b, c$ here refer to the coefficients of the standard quadratic equation. So, we need to find $A+B+C$. $A+B+C = 4 + (-9) + 0$ $A+B+C = 4 - 9$ $A+B+C = -5$

The condition $a \neq 0$ is given to ensure that $\sqrt{a}$ is well-defined and positive if 'a' is real and positive, and also to exclude $a=0$ as a solution if we were solving for 'a'. However, for forming the quadratic equation and finding the sum of its coefficients, this condition does not change the coefficients themselves. The quadratic equation $4a^2 - 9a = 0$ has solutions $a=0$ and $a=9/4$. When considering the original equation $2a=3\sqrt{a}$ and the condition $a \neq 0$, only $a=9/4$ is the valid solution. But the question is about the coefficients of the equation itself, not its solution.