+26367 x^2+3x+4=0 what does the x equal? please solve
\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0\\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac}} {2a} \\ \end{array} ~}\)
\(\begin{array}{rcll} x^2+3x+4 &=& 0 \qquad & a=1 \quad b=3 \quad c = 4 \\\\ x &=& \frac{-3 \pm \sqrt{3^2-4\cdot 1 \cdot 4} }{2\cdot 1} \\\\ x &=& \frac{-3 \pm \sqrt{9-16} } {2} \\\\ x &=& \frac{-3 \pm \sqrt{-7} } {2} \qquad &| \quad \sqrt{-7}=\sqrt{-1\cdot 7}=\sqrt{-1}\sqrt{7}\\\\ x &=& \frac{-3 \pm \sqrt{-1}\sqrt{7} } {2} \qquad &| \quad \sqrt{-1} = i \\\\ x &=& \frac{-3 \pm i\sqrt{7} } {2} \\\\ \mathbf{x_1} &\mathbf{=}& \mathbf{-\frac32 + \frac{\sqrt{7}}{2}\cdot i } \\\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-\frac32 - \frac{\sqrt{7}}{2}\cdot i } \end{array}\)
Plugging this into WolframAlpha gives me this (admittedly dubious) answer.
x = 1/2*i*(sqrt(7)+3i
x = 1/2*i*(sqrt(7)-3i
solve for x, can use quadratic formula, factoring, completing the square etc.
http://www.dummies.com/how-to/content/how-to-factor-a-polynomial-expression.html
doesn't factor, use quadratic formula or complete the square
completing the square:
x^2+3x+4 = 0 Note: ax^2+bx+c (a, b & c are constants not variables like x)
x^2+3x = -4
x^2+3x+2.25=-4+2.25 Note: (b/2)^2 = c 3/2 = 1.5, 1.5^2 = 2.25 (add 2.25 to both sides to keep the equation the same)
(x+1.5)^2 = 1.75 Note: x^2+3x+2.25 using complete the square is (x+1.5)^2
sqrt(x+1.5^2)= sqrt(1.75) Note: Again, same operation to both sides, square root to keep the equation the same
x+1.5 = (plus or minus)sqrt(1.75)
x = (plus or minus)sqrt(1.75) - 1.5
There can only be one guest
solve for x, can use quadratic formula, factoring, completing the square etc.
http://www.dummies.com/how-to/content/how-to-factor-a-polynomial-expression.html
doesn't factor, use quadratic formula or complete the square
completing the square:
x^2+3x+4 = 0 Note: ax^2+bx+c (a, b & c are constants not variables like x)
x^2+3x = -4
x^2+3x+2.25=-4+2.25 Note: (b/2)^2 = c 3/2 = 1.5, 1.5^2 = 2.25 (add 2.25 to both sides to keep the equation the same)
(x+1.5)^2 = -1.75 Note: x^2+3x+2.25 using complete the square is (x+1.5)^2
sqrt(x+1.5^2)= sqrt(-1.75) Note: Again, same operation to both sides, square root to keep the equation the same
x+1.5 = (plus or minus) sqrt(-1.75)
x = (plus or minus) sqrt(-1.75) - 1.5
carry negative sign* it does exist just not a real number it is an imaginary number
+26367 x^2+3x+4=0 what does the x equal? please solve
\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0\\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac}} {2a} \\ \end{array} ~}\)
\(\begin{array}{rcll} x^2+3x+4 &=& 0 \qquad & a=1 \quad b=3 \quad c = 4 \\\\ x &=& \frac{-3 \pm \sqrt{3^2-4\cdot 1 \cdot 4} }{2\cdot 1} \\\\ x &=& \frac{-3 \pm \sqrt{9-16} } {2} \\\\ x &=& \frac{-3 \pm \sqrt{-7} } {2} \qquad &| \quad \sqrt{-7}=\sqrt{-1\cdot 7}=\sqrt{-1}\sqrt{7}\\\\ x &=& \frac{-3 \pm \sqrt{-1}\sqrt{7} } {2} \qquad &| \quad \sqrt{-1} = i \\\\ x &=& \frac{-3 \pm i\sqrt{7} } {2} \\\\ \mathbf{x_1} &\mathbf{=}& \mathbf{-\frac32 + \frac{\sqrt{7}}{2}\cdot i } \\\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-\frac32 - \frac{\sqrt{7}}{2}\cdot i } \end{array}\)
