Question
Mathematics
What are the solutions of the equation $6{x}^{2}+x-1=0$6x^(2)+x-1=0 ?
A. $\frac{1}{2}$(1)/(2)
B. $-\frac{1}{3}$-(1)/(3)
C. $-\frac{1}{2}$-(1)/(2)
D. $\frac{1}{3}$(1)/(3)
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To solve the quadratic equation $6{x}^{2}+x-1=0$6x^(2)+x-1=0, we can either factor the equation or use the quadratic formula.
Using the quadratic formula: Given an equation of the form $a{x}^{2}+bx+c=0$ax^(2)+bx+c=0, the solutions for $x$x are given by:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}.$x=(-b+-sqrt(b^(2)-4ac))/(2a).
For our equation $6{x}^{2}+x-1=0$6x^(2)+x-1=0, we have $a=6$a=6, $b=1$b=1, and $c=-1$c=-1. Substituting these values into the quadratic formula:
$x=\frac{-1±\sqrt{{1}^{2}-4\left(6\right)\left(-1\right)}}{2\left(6\right)}.$x=(-1+-sqrt(1^(2)-4(6)(-1)))/(2(6)).
Simplifying further:
$x=\frac{-1±\sqrt{1+24}}{12}.$x=(-1+-sqrt(1+24))/(12).
$x=\frac{-1±\sqrt{25}}{12}.$x=(-1+-sqrt25)/(12).
$x=\frac{-1±5}{12}.$x=(-1+-5)/(12).
We have two solutions:
$x=\frac{-1+5}{12}=\frac{4}{12}=\frac{1}{3}\text{}\text{(Solution 1)}.$x=(-1+5)/(12)=(4)/(12)=(1)/(3)(Solution 1).
$x=\frac{-1-5}{12}=\frac{-6}{12}=-\frac{1}{2}\text{}\text{(Solution 2)}.$x=(-1-5)/(12)=(-6)/(12)=-(1)/(2)(Solution 2).
Therefore, the solutions of the equation $6{x}^{2}+x-1=0$6x^(2)+x-1=0 are $\overline{)\mathbf{\text{(D)}}\phantom{\rule{0ex}{0ex}}\frac{1}{3}}$(D)(1)/(3) and $\overline{)\mathbf{\text{(C)}}\phantom{\rule{0ex}{0ex}}-\frac{1}{2}}$(C)-(1)/(2).
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