Question
Solve the following quadratic equations by factorization:
$4x^2 + abx - (a^2 - b^2) = 0$
$4x^2 + abx - (a^2 - b^2) = 0$
✓
Answer
We have been given
$4x^2 + abx - (a^2 - b^2) = 0$
$4x^2 + 2(a+b)x - 2(a - b)x - (a^2 - b^2) = 0$
$2x(2x + a + b) - (a - b)(2x + a + b) = 0$
$(2x - (a - b))(2x + a + b) = 0$
Therefore,
$2x - (a - b) = 0$
$2x = a - b$
$\text{x}=\frac{\text{a}-\text{b}}{2}$
or, $2x + a + b = 0$
$2x = -(a + b)$
$\text{x}=\frac{-(\text{a}+\text{b})}{2}$
Hence, $\text{x}=\frac{\text{a}-\text{b}}{2}$ or $\text{x}=\frac{-(\text{a}+\text{b})}{2}$
$4x^2 + abx - (a^2 - b^2) = 0$
$4x^2 + 2(a+b)x - 2(a - b)x - (a^2 - b^2) = 0$
$2x(2x + a + b) - (a - b)(2x + a + b) = 0$
$(2x - (a - b))(2x + a + b) = 0$
Therefore,
$2x - (a - b) = 0$
$2x = a - b$
$\text{x}=\frac{\text{a}-\text{b}}{2}$
or, $2x + a + b = 0$
$2x = -(a + b)$
$\text{x}=\frac{-(\text{a}+\text{b})}{2}$
Hence, $\text{x}=\frac{\text{a}-\text{b}}{2}$ or $\text{x}=\frac{-(\text{a}+\text{b})}{2}$
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