Certainly! Let's solve the problem step-by-step.
We are given the equation:
[tex]\[
2a - b + c = 0
\][/tex]
We need to demonstrate that:
[tex]\[
4a^2 - b^2 + c^2 + 4ac = 0
\][/tex]
First, let's express [tex]\(b\)[/tex] in terms of [tex]\(a\)[/tex] and [tex]\(c\)[/tex] from the given equation [tex]\(2a - b + c = 0\)[/tex]:
[tex]\[
2a + c = b
\][/tex]
Now, we substitute [tex]\(b = 2a + c\)[/tex] into the expression we need to prove:
Consider the expression [tex]\(4a^2 - b^2 + c^2 + 4ac\)[/tex]:
By substituting [tex]\(b = 2a + c\)[/tex], the expression becomes:
[tex]\[
4a^2 - (2a + c)^2 + c^2 + 4ac
\][/tex]
Next, let's expand [tex]\((2a + c)^2\)[/tex]:
[tex]\[
(2a + c)^2 = 4a^2 + 4ac + c^2
\][/tex]
Now, substitute this expansion back into our expression:
[tex]\[
4a^2 - (4a^2 + 4ac + c^2) + c^2 + 4ac
\][/tex]
Simplify this step-by-step:
[tex]\[
= 4a^2 - 4a^2 - 4ac - c^2 + c^2 + 4ac
\][/tex]
Combine like terms:
[tex]\[
= 4a^2 - 4a^2 - 4ac - c^2 + c^2 + 4ac
\][/tex]
Notice that everything cancels out:
[tex]\[
= 0
\][/tex]
Thus, we have shown that:
[tex]\[
4a^2 - b^2 + c^2 + 4ac = 0
\][/tex]
Therefore, given [tex]\(2a - b + c = 0\)[/tex], we've proven that [tex]\(4a^2 - b^2 + c^2 + 4ac = 0\)[/tex].
