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a+b=29 ab=7\times 4=28
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7a^{2}+aa+ba+4. To find a and b, set up a system to be solved.
1,28 2,14 4,7
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 28.
1+28=29 2+14=16 4+7=11
Calculate the sum for each pair.
a=1 b=28
The solution is the pair that gives sum 29.
\left(7a^{2}+a\right)+\left(28a+4\right)
Rewrite 7a^{2}+29a+4 as \left(7a^{2}+a\right)+\left(28a+4\right).
a\left(7a+1\right)+4\left(7a+1\right)
Factor out a in the first and 4 in the second group.
\left(7a+1\right)\left(a+4\right)
Factor out common term 7a+1 by using distributive property.
a=-\frac{1}{7} a=-4
To find equation solutions, solve 7a+1=0 and a+4=0.
7a^{2}+29a+4=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.
a=\frac{-29±\sqrt{29^{2}-4\times 7\times 4}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 29 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-29±\sqrt{841-4\times 7\times 4}}{2\times 7}
Square 29.
a=\frac{-29±\sqrt{841-28\times 4}}{2\times 7}
Multiply -4 times 7.
a=\frac{-29±\sqrt{841-112}}{2\times 7}
Multiply -28 times 4.
a=\frac{-29±\sqrt{729}}{2\times 7}
Add 841 to -112.
a=\frac{-29±27}{2\times 7}
Take the square root of 729.
a=\frac{-29±27}{14}
Multiply 2 times 7.
a=-\frac{2}{14}
Now solve the equation a=\frac{-29±27}{14} when ± is plus. Add -29 to 27.
a=-\frac{1}{7}
Reduce the fraction \frac{-2}{14} to lowest terms by extracting and canceling out 2.
a=-\frac{56}{14}
Now solve the equation a=\frac{-29±27}{14} when ± is minus. Subtract 27 from -29.
a=-4
Divide -56 by 14.
a=-\frac{1}{7} a=-4
The equation is now solved.
7a^{2}+29a+4=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.
7a^{2}+29a+4-4=-4
Subtract 4 from both sides of the equation.
7a^{2}+29a=-4
Subtracting 4 from itself leaves 0.
\frac{7a^{2}+29a}{7}=-\frac{4}{7}
Divide both sides by 7.
a^{2}+\frac{29}{7}a=-\frac{4}{7}
Dividing by 7 undoes the multiplication by 7.
a^{2}+\frac{29}{7}a+\left(\frac{29}{14}\right)^{2}=-\frac{4}{7}+\left(\frac{29}{14}\right)^{2}
Divide \frac{29}{7}, the coefficient of the x term, by 2 to get \frac{29}{14}. Then add the square of \frac{29}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}+\frac{29}{7}a+\frac{841}{196}=-\frac{4}{7}+\frac{841}{196}
Square \frac{29}{14} by squaring both the numerator and the denominator of the fraction.
a^{2}+\frac{29}{7}a+\frac{841}{196}=\frac{729}{196}
Add -\frac{4}{7} to \frac{841}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a+\frac{29}{14}\right)^{2}=\frac{729}{196}
Factor a^{2}+\frac{29}{7}a+\frac{841}{196}. 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(a+\frac{29}{14}\right)^{2}}=\sqrt{\frac{729}{196}}
Take the square root of both sides of the equation.
a+\frac{29}{14}=\frac{27}{14} a+\frac{29}{14}=-\frac{27}{14}
Simplify.
a=-\frac{1}{7} a=-4
Subtract \frac{29}{14} from both sides of the equation.
x ^ 2 +\frac{29}{7}x +\frac{4}{7} = 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 7
r + s = -\frac{29}{7} rs = \frac{4}{7}
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{29}{14} - u s = -\frac{29}{14} + u
Two numbers r and s sum up to -\frac{29}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{29}{7} = -\frac{29}{14}. 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{29}{14} - u) (-\frac{29}{14} + u) = \frac{4}{7}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{7}
\frac{841}{196} - u^2 = \frac{4}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{4}{7}-\frac{841}{196} = -\frac{729}{196}
Simplify the expression by subtracting \frac{841}{196} on both sides
u^2 = \frac{729}{196} u = \pm\sqrt{\frac{729}{196}} = \pm \frac{27}{14}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{29}{14} - \frac{27}{14} = -4 s = -\frac{29}{14} + \frac{27}{14} = -0.143
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.