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a+b=7 ab=6\times 1=6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6d^{2}+ad+bd+1. To find a and b, set up a system to be solved.
1,6 2,3
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.
1+6=7 2+3=5
Calculate the sum for each pair.
a=1 b=6
The solution is the pair that gives sum 7.
\left(6d^{2}+d\right)+\left(6d+1\right)
Rewrite 6d^{2}+7d+1 as \left(6d^{2}+d\right)+\left(6d+1\right).
d\left(6d+1\right)+6d+1
Factor out d in 6d^{2}+d.
\left(6d+1\right)\left(d+1\right)
Factor out common term 6d+1 by using distributive property.
d=-\frac{1}{6} d=-1
To find equation solutions, solve 6d+1=0 and d+1=0.
6d^{2}+7d+1=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.
d=\frac{-7±\sqrt{7^{2}-4\times 6}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 7 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
d=\frac{-7±\sqrt{49-4\times 6}}{2\times 6}
Square 7.
d=\frac{-7±\sqrt{49-24}}{2\times 6}
Multiply -4 times 6.
d=\frac{-7±\sqrt{25}}{2\times 6}
Add 49 to -24.
d=\frac{-7±5}{2\times 6}
Take the square root of 25.
d=\frac{-7±5}{12}
Multiply 2 times 6.
d=-\frac{2}{12}
Now solve the equation d=\frac{-7±5}{12} when ± is plus. Add -7 to 5.
d=-\frac{1}{6}
Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.
d=-\frac{12}{12}
Now solve the equation d=\frac{-7±5}{12} when ± is minus. Subtract 5 from -7.
d=-1
Divide -12 by 12.
d=-\frac{1}{6} d=-1
The equation is now solved.
6d^{2}+7d+1=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.
6d^{2}+7d+1-1=-1
Subtract 1 from both sides of the equation.
6d^{2}+7d=-1
Subtracting 1 from itself leaves 0.
\frac{6d^{2}+7d}{6}=-\frac{1}{6}
Divide both sides by 6.
d^{2}+\frac{7}{6}d=-\frac{1}{6}
Dividing by 6 undoes the multiplication by 6.
d^{2}+\frac{7}{6}d+\left(\frac{7}{12}\right)^{2}=-\frac{1}{6}+\left(\frac{7}{12}\right)^{2}
Divide \frac{7}{6}, the coefficient of the x term, by 2 to get \frac{7}{12}. Then add the square of \frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
d^{2}+\frac{7}{6}d+\frac{49}{144}=-\frac{1}{6}+\frac{49}{144}
Square \frac{7}{12} by squaring both the numerator and the denominator of the fraction.
d^{2}+\frac{7}{6}d+\frac{49}{144}=\frac{25}{144}
Add -\frac{1}{6} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(d+\frac{7}{12}\right)^{2}=\frac{25}{144}
Factor d^{2}+\frac{7}{6}d+\frac{49}{144}. 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(d+\frac{7}{12}\right)^{2}}=\sqrt{\frac{25}{144}}
Take the square root of both sides of the equation.
d+\frac{7}{12}=\frac{5}{12} d+\frac{7}{12}=-\frac{5}{12}
Simplify.
d=-\frac{1}{6} d=-1
Subtract \frac{7}{12} from both sides of the equation.
x ^ 2 +\frac{7}{6}x +\frac{1}{6} = 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 6
r + s = -\frac{7}{6} rs = \frac{1}{6}
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 = -\frac{7}{12} - u s = -\frac{7}{12} + u
Two numbers r and s sum up to -\frac{7}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{6} = -\frac{7}{12}. 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>
(-\frac{7}{12} - u) (-\frac{7}{12} + u) = \frac{1}{6}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{6}
\frac{49}{144} - u^2 = \frac{1}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{6}-\frac{49}{144} = -\frac{25}{144}
Simplify the expression by subtracting \frac{49}{144} on both sides
u^2 = \frac{25}{144} u = \pm\sqrt{\frac{25}{144}} = \pm \frac{5}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{7}{12} - \frac{5}{12} = -1 s = -\frac{7}{12} + \frac{5}{12} = -0.167
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.