Find the roots of the following quadratic equations by using the quadratic formula $\displaystyle \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} = 6\frac{6}{7}, x \neq - 3, 3$
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Find the roots of the following quadratic equations by using the quadratic formula $\displaystyle \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} = 6\frac{6}{7}, x \neq - 3, 3$
Let y = (x-3)/(x+3). Equation: y - 1/y = 48/7. (y^2 - 1)/y = 48/7. 7y^2 - 48y - 7 = 0. (7y + 1)(y - 7) = 0. y = 7 or y = -1/7. If (x-3)/(x+3) = 7, x-3 = 7x+21, -6x = 24, x = -4. If (x-3)/(x+3) = -1/7, 7x-21 = -x-3, 8x = 18, x = 9/4.
Multiply the equation by the least common denominator, which is the quantity x plus 3 times the quantity x minus 3, to clear the fractions and get the quantity x minus 3 squared minus the quantity x plus 3 squared equals the fraction 48 over 7 times the quantity x squared minus 9. Expanding both sides yields negative 12x equals the fraction 48 over 7 times x squared minus 9, which rearranges to the quadratic equation 4x squared plus 7x minus 36 equals 0. Applying the quadratic formula with a equals 4, b equals 7, and c equals negative 36 gives x equals negative 4 or x equals 9 fourths.