\( \frac { 7 \pm \sqrt { x ^ { 2 } + 3 x - 4 } = 0 } { x ^ { 2 } } \)
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To solve the equation \( \frac{7 \pm \sqrt{x^2 + 3x - 4}}{x^2} = 0 \), we first need to recognize that the fraction can only equal zero if the numerator is zero (as long as the denominator isn't). This means we need to set the numerator \( 7 \pm \sqrt{x^2 + 3x - 4} = 0 \). For the positive case \( 7 + \sqrt{x^2 + 3x - 4} = 0 \), this can't happen because \( \sqrt{x^2 + 3x - 4} \) is always non-negative, thus no solution from this case. Now for the negative case \( 7 - \sqrt{x^2 + 3x - 4} = 0 \), we can simplify it to \( \sqrt{x^2 + 3x - 4} = 7 \). Squaring both sides gives us \( x^2 + 3x - 4 = 49 \), leading us to \( x^2 + 3x - 53 = 0 \). Now, applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=3, c=-53 \): \( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-53)}}{2(1)} \) \( x = \frac{-3 \pm \sqrt{9 + 212}}{2} \) \( x = \frac{-3 \pm \sqrt{221}}{2} \) So, the final solutions are \( x = \frac{-3 + \sqrt{221}}{2} \) and \( x = \frac{-3 - \sqrt{221}}{2} \). This gives us a methodical approach to tackle equations involving square roots and fractions. Always check for extraneous solutions, especially when squaring both sides!