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3x^{2}+24x+49=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-24±\sqrt{24^{2}-4\times 3\times 49}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 24 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-24±\sqrt{576-4\times 3\times 49}}{2\times 3}
Square 24.
x=\frac{-24±\sqrt{576-12\times 49}}{2\times 3}
Multiply -4 times 3.
x=\frac{-24±\sqrt{576-588}}{2\times 3}
Multiply -12 times 49.
x=\frac{-24±\sqrt{-12}}{2\times 3}
Add 576 to -588.
x=\frac{-24±2\sqrt{3}i}{2\times 3}
Take the square root of -12.
x=\frac{-24±2\sqrt{3}i}{6}
Multiply 2 times 3.
x=\frac{-24+2\sqrt{3}i}{6}
Now solve the equation x=\frac{-24±2\sqrt{3}i}{6} when ± is plus. Add -24 to 2i\sqrt{3}.
x=\frac{\sqrt{3}i}{3}-4
Divide -24+2i\sqrt{3} by 6.
x=\frac{-2\sqrt{3}i-24}{6}
Now solve the equation x=\frac{-24±2\sqrt{3}i}{6} when ± is minus. Subtract 2i\sqrt{3} from -24.
x=-\frac{\sqrt{3}i}{3}-4
Divide -24-2i\sqrt{3} by 6.
x=\frac{\sqrt{3}i}{3}-4 x=-\frac{\sqrt{3}i}{3}-4
The equation is now solved.
3x^{2}+24x+49=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
3x^{2}+24x+49-49=-49
Subtract 49 from both sides of the equation.
3x^{2}+24x=-49
Subtracting 49 from itself leaves 0.
\frac{3x^{2}+24x}{3}=-\frac{49}{3}
Divide both sides by 3.
x^{2}+\frac{24}{3}x=-\frac{49}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+8x=-\frac{49}{3}
Divide 24 by 3.
x^{2}+8x+4^{2}=-\frac{49}{3}+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=-\frac{49}{3}+16
Square 4.
x^{2}+8x+16=-\frac{1}{3}
Add -\frac{49}{3} to 16.
\left(x+4\right)^{2}=-\frac{1}{3}
Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+4\right)^{2}}=\sqrt{-\frac{1}{3}}
Take the square root of both sides of the equation.
x+4=\frac{\sqrt{3}i}{3} x+4=-\frac{\sqrt{3}i}{3}
Simplify.
x=\frac{\sqrt{3}i}{3}-4 x=-\frac{\sqrt{3}i}{3}-4
Subtract 4 from both sides of the equation.
x ^ 2 +8x +\frac{49}{3} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3
r + s = -8 rs = \frac{49}{3}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -4 - u s = -4 + u
Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-4 - u) (-4 + u) = \frac{49}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{49}{3}
16 - u^2 = \frac{49}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{49}{3}-16 = \frac{1}{3}
Simplify the expression by subtracting 16 on both sides
u^2 = -\frac{1}{3} u = \pm\sqrt{-\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-4 - \frac{1}{\sqrt{3}}i = -4 - 0.577i s = -4 + \frac{1}{\sqrt{3}}i = -4 + 0.577i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.