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What are the solutions of the equation 6 x 2 + x 1 = 0 6x^(2)+x-1=0 ?
A. 1 2 (1)/(2)
B. 1 3 -(1)/(3)
C. 1 2 -(1)/(2)
D. 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 + b x + c = 0 ax^(2)+bx+c=0, the solutions for x x are given by:
x = b ± b 2 4 a c 2 a . 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 = 1 ± 1 2 4 ( 6 ) ( 1 ) 2 ( 6 ) . x=(-1+-sqrt(1^(2)-4(6)(-1)))/(2(6)).
Simplifying further:
x = 1 ± 1 + 24 12 . x=(-1+-sqrt(1+24))/(12).
x = 1 ± 25 12 . x=(-1+-sqrt25)/(12).
x = 1 ± 5 12 . x=(-1+-5)/(12).
We have two solutions:
x = 1 + 5 12 = 4 12 = 1 3 (Solution 1) . x=(-1+5)/(12)=(4)/(12)=(1)/(3)(Solution 1).
x = 1 5 12 = 6 12 = 1 2 (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 (D) 1 3 (D)(1)/(3) and (C) 1 2 (C)-(1)/(2).
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