Solve for x
x=-\frac{1}{2}=-0.5
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a+b=4 ab=4\times 1=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+1. To find a and b, set up a system to be solved.
1,4 2,2
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 4.
1+4=5 2+2=4
Calculate the sum for each pair.
a=2 b=2
The solution is the pair that gives sum 4.
\left(4x^{2}+2x\right)+\left(2x+1\right)
Rewrite 4x^{2}+4x+1 as \left(4x^{2}+2x\right)+\left(2x+1\right).
2x\left(2x+1\right)+2x+1
Factor out 2x in 4x^{2}+2x.
\left(2x+1\right)\left(2x+1\right)
Factor out common term 2x+1 by using distributive property.
\left(2x+1\right)^{2}
Rewrite as a binomial square.
x=-\frac{1}{2}
To find equation solution, solve 2x+1=0.
4x^{2}+4x+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.
x=\frac{-4±\sqrt{4^{2}-4\times 4}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\times 4}}{2\times 4}
Square 4.
x=\frac{-4±\sqrt{16-16}}{2\times 4}
Multiply -4 times 4.
x=\frac{-4±\sqrt{0}}{2\times 4}
Add 16 to -16.
x=-\frac{4}{2\times 4}
Take the square root of 0.
x=-\frac{4}{8}
Multiply 2 times 4.
x=-\frac{1}{2}
Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.
4x^{2}+4x+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.
4x^{2}+4x+1-1=-1
Subtract 1 from both sides of the equation.
4x^{2}+4x=-1
Subtracting 1 from itself leaves 0.
\frac{4x^{2}+4x}{4}=-\frac{1}{4}
Divide both sides by 4.
x^{2}+\frac{4}{4}x=-\frac{1}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+x=-\frac{1}{4}
Divide 4 by 4.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{1}{4}+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+x+\frac{1}{4}=\frac{-1+1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=0
Add -\frac{1}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{2}\right)^{2}=0
Factor x^{2}+x+\frac{1}{4}. 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{1}{2}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x+\frac{1}{2}=0 x+\frac{1}{2}=0
Simplify.
x=-\frac{1}{2} x=-\frac{1}{2}
Subtract \frac{1}{2} from both sides of the equation.
x=-\frac{1}{2}
The equation is now solved. Solutions are the same.
x ^ 2 +1x +\frac{1}{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 = -1 rs = \frac{1}{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{1}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. 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{1}{2} - u) (-\frac{1}{2} + u) = \frac{1}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{4}
\frac{1}{4} - u^2 = \frac{1}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{4}-\frac{1}{4} = 0
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = -\frac{1}{2} = -0.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ 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}