Solve for x
x=-4
x=-\frac{1}{5}=-0.2
Graph
Share
Copied to clipboard
a+b=21 ab=5\times 4=20
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+4. To find a and b, set up a system to be solved.
1,20 2,10 4,5
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 20.
1+20=21 2+10=12 4+5=9
Calculate the sum for each pair.
a=1 b=20
The solution is the pair that gives sum 21.
\left(5x^{2}+x\right)+\left(20x+4\right)
Rewrite 5x^{2}+21x+4 as \left(5x^{2}+x\right)+\left(20x+4\right).
x\left(5x+1\right)+4\left(5x+1\right)
Factor out x in the first and 4 in the second group.
\left(5x+1\right)\left(x+4\right)
Factor out common term 5x+1 by using distributive property.
x=-\frac{1}{5} x=-4
To find equation solutions, solve 5x+1=0 and x+4=0.
5x^{2}+21x+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.
x=\frac{-21±\sqrt{21^{2}-4\times 5\times 4}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 21 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-21±\sqrt{441-4\times 5\times 4}}{2\times 5}
Square 21.
x=\frac{-21±\sqrt{441-20\times 4}}{2\times 5}
Multiply -4 times 5.
x=\frac{-21±\sqrt{441-80}}{2\times 5}
Multiply -20 times 4.
x=\frac{-21±\sqrt{361}}{2\times 5}
Add 441 to -80.
x=\frac{-21±19}{2\times 5}
Take the square root of 361.
x=\frac{-21±19}{10}
Multiply 2 times 5.
x=-\frac{2}{10}
Now solve the equation x=\frac{-21±19}{10} when ± is plus. Add -21 to 19.
x=-\frac{1}{5}
Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.
x=-\frac{40}{10}
Now solve the equation x=\frac{-21±19}{10} when ± is minus. Subtract 19 from -21.
x=-4
Divide -40 by 10.
x=-\frac{1}{5} x=-4
The equation is now solved.
5x^{2}+21x+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.
5x^{2}+21x+4-4=-4
Subtract 4 from both sides of the equation.
5x^{2}+21x=-4
Subtracting 4 from itself leaves 0.
\frac{5x^{2}+21x}{5}=-\frac{4}{5}
Divide both sides by 5.
x^{2}+\frac{21}{5}x=-\frac{4}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{21}{5}x+\left(\frac{21}{10}\right)^{2}=-\frac{4}{5}+\left(\frac{21}{10}\right)^{2}
Divide \frac{21}{5}, the coefficient of the x term, by 2 to get \frac{21}{10}. Then add the square of \frac{21}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{21}{5}x+\frac{441}{100}=-\frac{4}{5}+\frac{441}{100}
Square \frac{21}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{21}{5}x+\frac{441}{100}=\frac{361}{100}
Add -\frac{4}{5} to \frac{441}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{21}{10}\right)^{2}=\frac{361}{100}
Factor x^{2}+\frac{21}{5}x+\frac{441}{100}. 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+\frac{21}{10}\right)^{2}}=\sqrt{\frac{361}{100}}
Take the square root of both sides of the equation.
x+\frac{21}{10}=\frac{19}{10} x+\frac{21}{10}=-\frac{19}{10}
Simplify.
x=-\frac{1}{5} x=-4
Subtract \frac{21}{10} from both sides of the equation.
x ^ 2 +\frac{21}{5}x +\frac{4}{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{21}{5} rs = \frac{4}{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{21}{10} - u s = -\frac{21}{10} + u
Two numbers r and s sum up to -\frac{21}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{21}{5} = -\frac{21}{10}. 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{21}{10} - u) (-\frac{21}{10} + u) = \frac{4}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{5}
\frac{441}{100} - u^2 = \frac{4}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{4}{5}-\frac{441}{100} = -\frac{361}{100}
Simplify the expression by subtracting \frac{441}{100} on both sides
u^2 = \frac{361}{100} u = \pm\sqrt{\frac{361}{100}} = \pm \frac{19}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{21}{10} - \frac{19}{10} = -4 s = -\frac{21}{10} + \frac{19}{10} = -0.200
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}