Solve for a
a=\frac{2}{5}=0.4
a=4
Share
Copied to clipboard
a+b=-22 ab=5\times 8=40
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5a^{2}+aa+ba+8. To find a and b, set up a system to be solved.
-1,-40 -2,-20 -4,-10 -5,-8
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 40.
-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13
Calculate the sum for each pair.
a=-20 b=-2
The solution is the pair that gives sum -22.
\left(5a^{2}-20a\right)+\left(-2a+8\right)
Rewrite 5a^{2}-22a+8 as \left(5a^{2}-20a\right)+\left(-2a+8\right).
5a\left(a-4\right)-2\left(a-4\right)
Factor out 5a in the first and -2 in the second group.
\left(a-4\right)\left(5a-2\right)
Factor out common term a-4 by using distributive property.
a=4 a=\frac{2}{5}
To find equation solutions, solve a-4=0 and 5a-2=0.
5a^{2}-22a+8=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{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 5\times 8}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -22 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-22\right)±\sqrt{484-4\times 5\times 8}}{2\times 5}
Square -22.
a=\frac{-\left(-22\right)±\sqrt{484-20\times 8}}{2\times 5}
Multiply -4 times 5.
a=\frac{-\left(-22\right)±\sqrt{484-160}}{2\times 5}
Multiply -20 times 8.
a=\frac{-\left(-22\right)±\sqrt{324}}{2\times 5}
Add 484 to -160.
a=\frac{-\left(-22\right)±18}{2\times 5}
Take the square root of 324.
a=\frac{22±18}{2\times 5}
The opposite of -22 is 22.
a=\frac{22±18}{10}
Multiply 2 times 5.
a=\frac{40}{10}
Now solve the equation a=\frac{22±18}{10} when ± is plus. Add 22 to 18.
a=4
Divide 40 by 10.
a=\frac{4}{10}
Now solve the equation a=\frac{22±18}{10} when ± is minus. Subtract 18 from 22.
a=\frac{2}{5}
Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.
a=4 a=\frac{2}{5}
The equation is now solved.
5a^{2}-22a+8=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.
5a^{2}-22a+8-8=-8
Subtract 8 from both sides of the equation.
5a^{2}-22a=-8
Subtracting 8 from itself leaves 0.
\frac{5a^{2}-22a}{5}=-\frac{8}{5}
Divide both sides by 5.
a^{2}-\frac{22}{5}a=-\frac{8}{5}
Dividing by 5 undoes the multiplication by 5.
a^{2}-\frac{22}{5}a+\left(-\frac{11}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{11}{5}\right)^{2}
Divide -\frac{22}{5}, the coefficient of the x term, by 2 to get -\frac{11}{5}. Then add the square of -\frac{11}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{22}{5}a+\frac{121}{25}=-\frac{8}{5}+\frac{121}{25}
Square -\frac{11}{5} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{22}{5}a+\frac{121}{25}=\frac{81}{25}
Add -\frac{8}{5} to \frac{121}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{11}{5}\right)^{2}=\frac{81}{25}
Factor a^{2}-\frac{22}{5}a+\frac{121}{25}. 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{11}{5}\right)^{2}}=\sqrt{\frac{81}{25}}
Take the square root of both sides of the equation.
a-\frac{11}{5}=\frac{9}{5} a-\frac{11}{5}=-\frac{9}{5}
Simplify.
a=4 a=\frac{2}{5}
Add \frac{11}{5} to both sides of the equation.
x ^ 2 -\frac{22}{5}x +\frac{8}{5} = 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 5
r + s = \frac{22}{5} rs = \frac{8}{5}
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{11}{5} - u s = \frac{11}{5} + u
Two numbers r and s sum up to \frac{22}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{22}{5} = \frac{11}{5}. 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{11}{5} - u) (\frac{11}{5} + u) = \frac{8}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{5}
\frac{121}{25} - u^2 = \frac{8}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{8}{5}-\frac{121}{25} = -\frac{81}{25}
Simplify the expression by subtracting \frac{121}{25} on both sides
u^2 = \frac{81}{25} u = \pm\sqrt{\frac{81}{25}} = \pm \frac{9}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{11}{5} - \frac{9}{5} = 0.400 s = \frac{11}{5} + \frac{9}{5} = 4
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}