What values of a and b make the equation true √648 = √(2a + 3b)?

To solve the equation √648 = √(2a + 3b), we first need to simplify the left side.

We can express 648 as a product of its prime factors:

648 = 2^3 * 3^4

This means:

√648 = √(2^3 * 3^4) = √(2^2 * 3^4 * 2) = 2 * 3^2 * √2 = 18√2.

So, to make our equation true:

√(2a + 3b) must equal 18√2.

Squaring both sides gives:

2a + 3b = (18√2)² = 324 * 2 = 648.

Now we are looking for values of a and b that satisfy:

2a + 3b = 648.

There are infinitely many combinations of (a, b) that can satisfy this equation. For example:

  • If a = 0, then 3b = 648b = 216.
  • If a = 324, then 3b = 0b = 0.

In conclusion, the values of a and b can vary as long as they satisfy the equation 2a + 3b = 648.

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