In solving the equation (x + 3)(x + 1) = 48, Eric stated that the solution would be
x + 3 = 48 => x = 45
or
(x + 1) = 48 => x = 47
However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem yourself, and explain your reasoning.
It doesn't work because if we assume:
x+3 = 48
we fail to take into account the x+1 expression.
However, if we had the equation (x+3)(x+1) = 0, then we can set x+3 = 0
It doesn't work because
ab = 48 does not mean that one or the other is 48.
However, if ab=0, then one or the other must be 0.
To solve the equation (x + 3)(x + 1) = 48, let's go through the process step by step:
1. Start by expanding the left side of the equation using the distributive property:
(x + 3)(x + 1) = x(x + 1) + 3(x + 1) = x^2 + x + 3x + 3 = x^2 + 4x + 3
2. Now, rewrite the equation with the expanded form:
x^2 + 4x + 3 = 48
3. Simplify the equation by moving all terms to one side:
x^2 + 4x + 3 - 48 = 0
4. Combine like terms:
x^2 + 4x - 45 = 0
5. Now, let's factor the quadratic equation. We are looking for two numbers that multiply to -45 and add up to 4. The numbers are 9 and -5:
(x + 9)(x - 5) = 0
6. Apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero:
x + 9 = 0 or x - 5 = 0
7. Solve the individual equations:
For x + 9 = 0, subtract 9 from both sides to isolate x: x = -9
For x - 5 = 0, add 5 to both sides to isolate x: x = 5
Therefore, the solutions for the equation (x + 3)(x + 1) = 48 are x = -9 and x = 5.
Now, let's check if these solutions work by substituting them back into the original equation:
For x = -9:
(x + 3)(x + 1) = (-9 + 3)(-9 + 1) = (-6)(-8) = 48
For x = 5:
(x + 3)(x + 1) = (5 + 3)(5 + 1) = (8)(6) = 48
Both solutions match the original equation, so both x = -9 and x = 5 are valid solutions. It seems that Eric's calculations may have been incorrect, which is why one of his solutions did not work when substituted back into the original equation.