The roots of the equation $\displaystyle 2{ y }^{ 2 }+y-2=0$ are
- $\displaystyle \frac { -1-\sqrt { 17 } }{ 2 } ,\frac { -1-\sqrt { 17 } }{ 2 } $
- $\displaystyle \frac {- 1-\sqrt { 17 } }{ 4 } ,\frac { -1+\sqrt { 17 } }{ 4 } $
- $\displaystyle -1-\sqrt { 17 } ,-1+\sqrt { 17 } $
-
None
For 2y^2 + y - 2 = 0, use the quadratic formula y = (-b +/- sqrt(b^2 - 4ac)) / 2a. Here a=2, b=1, c=-2. y = (-1 +/- sqrt(1 - 4*2*(-2))) / 4 = (-1 +/- sqrt(17)) / 4.
Use the quadratic formula to find the roots, which states y equals the quantity negative b plus or minus the square root of b squared minus 4ac all divided by 2a. Substituting a equals 2, b equals 1, and c equals negative 2 into the formula gives y equals negative 1 plus or minus the square root of 1 squared minus 4 times 2 times negative 2, all divided by 4. This simplifies to y equals negative 1 plus or minus the square root of 17 divided by 4, resulting in the roots the quantity negative 1 minus the square root of 17 divided by 4 and the quantity negative 1 plus the square root of 17 divided by 4.