Solve for s
s=-7
s=\frac{1}{4}=0.25
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a+b=27 ab=4\left(-7\right)=-28
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4s^{2}+as+bs-7. To find a and b, set up a system to be solved.
-1,28 -2,14 -4,7
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -28.
-1+28=27 -2+14=12 -4+7=3
Calculate the sum for each pair.
a=-1 b=28
The solution is the pair that gives sum 27.
\left(4s^{2}-s\right)+\left(28s-7\right)
Rewrite 4s^{2}+27s-7 as \left(4s^{2}-s\right)+\left(28s-7\right).
s\left(4s-1\right)+7\left(4s-1\right)
Factor out s in the first and 7 in the second group.
\left(4s-1\right)\left(s+7\right)
Factor out common term 4s-1 by using distributive property.
s=\frac{1}{4} s=-7
To find equation solutions, solve 4s-1=0 and s+7=0.
4s^{2}+27s-7=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.
s=\frac{-27±\sqrt{27^{2}-4\times 4\left(-7\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 27 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-27±\sqrt{729-4\times 4\left(-7\right)}}{2\times 4}
Square 27.
s=\frac{-27±\sqrt{729-16\left(-7\right)}}{2\times 4}
Multiply -4 times 4.
s=\frac{-27±\sqrt{729+112}}{2\times 4}
Multiply -16 times -7.
s=\frac{-27±\sqrt{841}}{2\times 4}
Add 729 to 112.
s=\frac{-27±29}{2\times 4}
Take the square root of 841.
s=\frac{-27±29}{8}
Multiply 2 times 4.
s=\frac{2}{8}
Now solve the equation s=\frac{-27±29}{8} when ± is plus. Add -27 to 29.
s=\frac{1}{4}
Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.
s=-\frac{56}{8}
Now solve the equation s=\frac{-27±29}{8} when ± is minus. Subtract 29 from -27.
s=-7
Divide -56 by 8.
s=\frac{1}{4} s=-7
The equation is now solved.
4s^{2}+27s-7=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.
4s^{2}+27s-7-\left(-7\right)=-\left(-7\right)
Add 7 to both sides of the equation.
4s^{2}+27s=-\left(-7\right)
Subtracting -7 from itself leaves 0.
4s^{2}+27s=7
Subtract -7 from 0.
\frac{4s^{2}+27s}{4}=\frac{7}{4}
Divide both sides by 4.
s^{2}+\frac{27}{4}s=\frac{7}{4}
Dividing by 4 undoes the multiplication by 4.
s^{2}+\frac{27}{4}s+\left(\frac{27}{8}\right)^{2}=\frac{7}{4}+\left(\frac{27}{8}\right)^{2}
Divide \frac{27}{4}, the coefficient of the x term, by 2 to get \frac{27}{8}. Then add the square of \frac{27}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
s^{2}+\frac{27}{4}s+\frac{729}{64}=\frac{7}{4}+\frac{729}{64}
Square \frac{27}{8} by squaring both the numerator and the denominator of the fraction.
s^{2}+\frac{27}{4}s+\frac{729}{64}=\frac{841}{64}
Add \frac{7}{4} to \frac{729}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(s+\frac{27}{8}\right)^{2}=\frac{841}{64}
Factor s^{2}+\frac{27}{4}s+\frac{729}{64}. 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(s+\frac{27}{8}\right)^{2}}=\sqrt{\frac{841}{64}}
Take the square root of both sides of the equation.
s+\frac{27}{8}=\frac{29}{8} s+\frac{27}{8}=-\frac{29}{8}
Simplify.
s=\frac{1}{4} s=-7
Subtract \frac{27}{8} from both sides of the equation.
x ^ 2 +\frac{27}{4}x -\frac{7}{4} = 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 4
r + s = -\frac{27}{4} rs = -\frac{7}{4}
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{27}{8} - u s = -\frac{27}{8} + u
Two numbers r and s sum up to -\frac{27}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{27}{4} = -\frac{27}{8}. 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{27}{8} - u) (-\frac{27}{8} + u) = -\frac{7}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{4}
\frac{729}{64} - u^2 = -\frac{7}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{7}{4}-\frac{729}{64} = -\frac{841}{64}
Simplify the expression by subtracting \frac{729}{64} on both sides
u^2 = \frac{841}{64} u = \pm\sqrt{\frac{841}{64}} = \pm \frac{29}{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{27}{8} - \frac{29}{8} = -7 s = -\frac{27}{8} + \frac{29}{8} = 0.250
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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