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16. Use the quadratic formula to solve 6x^2-2x-4=0 -(2)/(3),1 (2)/(3),1 -(2)/(3),-1 (2)/(3),-1

Pergunta

16. Use the quadratic formula to solve 6x^2-2x-4=0
 -(2)/(3),1 
 (2)/(3),1 
 -(2)/(3),-1 
 (2)/(3),-1

16. Use the quadratic formula to solve 6x^2-2x-4=0 -(2)/(3),1 (2)/(3),1 -(2)/(3),-1 (2)/(3),-1

Solução

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JoanaMestre · Tutor por 5 anos

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To solve the quadratic equation $6x^{2}-2x-4=0$ using the quadratic formula, we can use the formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a$, $b$, and $c$ are the coefficients of the quadratic equation.<br /><br />In this case, $a = 6$, $b = -2$, and $c = -4$.<br /><br />Substituting these values into the quadratic formula, we get:<br /><br />$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(6)(-4)}}{2(6)}$<br /><br />Simplifying the expression inside the square root:<br /><br />$x = \frac{2 \pm \sqrt{4 + 96}}{12}$<br /><br />$x = \frac{2 \pm \sqrt{100}}{12}$<br /><br />$x = \frac{2 \pm 10}{12}$<br /><br />Therefore, the solutions to the quadratic equation $6x^{2}-2x-4=0$ are:<br /><br />$x = \frac{2 + 10}{12} = \frac{12}{12} = 1$<br /><br />$x = \frac{2 - 10}{12} = \frac{-8}{12} = -\frac{2}{3}$<br /><br />So, the correct answer is $\{ -\frac {2}{3},1\} $.
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