Solve the equation. x2+13x+2=0

Question

Solve the equation.   x2+13x+2=0

pseudoenergy34 · Accepted Answer

Step 1  The given equation is x2+13x+2=0  Step 2  x2(3)+13x(3)+2(3)=0(3)  3x2+x+6=0  x=−1±12−4⋅3⋅62⋅3  =−1±−716  =−1±i716  x=−1+i716,x=−1−i716  Thus, the solution of the equation is x=−1+i716,x=−1−i716

Geraldine Flores · Accepted Answer

Step 1: Combine the multiplied terms into one fraction

x^{2} + \frac{1}{3} x + 2 = 0

x^{2} + \frac{1 x}{3} + 2 = 0

Step 2: Multiply by 1

x^{2} + \frac{x}{3} + 2 = 0

Step 3: Find the common denominator

\frac{3 x^{2}}{3} + \frac{x}{3} + \frac{3 \cdot 2}{3} = 0

Step 4: Add fractions with a common denominator

\frac{3 x^{2} + x + 3 \cdot 2}{3} = 0

Step 5: Multiply the numbers

\frac{3 x^{2} + x + 6}{3} = 0

Step 6: Multiply all terms by the same number to get rid of the denominators

3 (\frac{3 x^{2} + x + 6}{3}) = 0

Step 7: Reduce the multiplied terms in the denominator

3 x^{2} + x + 6 = 0

Step 8: Use the formula for the roots of the quadratic equation

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

In the standard form, determine a, b and c from the original equation and insert them into the formula for the roots of the quadratic equation.

3 x^{2} + x + 6 = 0

a=3
b=1
c=6

x = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 3 \cdot 6}}{2 \cdot 3}

Step 9: Simplification

x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 3 \cdot 6}}{2 \cdot 3}

x = \frac{- 1 \pm \sqrt{1 - 72}}{2 \cdot 3}

x = \frac{- 1 \pm \sqrt{- 71}}{2 \cdot 3}

x = \frac{- 1 \pm \sqrt{- 71}}{6}

Step 10: There are no real solutions because the discriminant is negative
The square root of a negative number is not a real number
d=-71
Result: There is no solution

Solve the equation. x^{2}+\frac{1}{3}x+2=0

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