Solve for x

x=1<br/>x=4

$x=1$

$x=4$

$x=4$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Direct Factoring Method

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a+b=-5 ab=4

To solve the equation, factor x^{2}-5x+4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). 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 negative, a and b are both negative. List all such integer pairs that give product 4.

-1-4=-5 -2-2=-4

Calculate the sum for each pair.

a=-4 b=-1

The solution is the pair that gives sum -5.

\left(x-4\right)\left(x-1\right)

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

x=4 x=1

To find equation solutions, solve x-4=0 and x-1=0.

a+b=-5 ab=1\times 4=4

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. 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 negative, a and b are both negative. List all such integer pairs that give product 4.

-1-4=-5 -2-2=-4

Calculate the sum for each pair.

a=-4 b=-1

The solution is the pair that gives sum -5.

\left(x^{2}-4x\right)+\left(-x+4\right)

Rewrite x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).

x\left(x-4\right)-\left(x-4\right)

Factor out x in the first and -1 in the second group.

\left(x-4\right)\left(x-1\right)

Factor out common term x-4 by using distributive property.

x=4 x=1

To find equation solutions, solve x-4=0 and x-1=0.

x^{2}-5x+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{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2}

Square -5.

x=\frac{-\left(-5\right)±\sqrt{25-16}}{2}

Multiply -4 times 4.

x=\frac{-\left(-5\right)±\sqrt{9}}{2}

Add 25 to -16.

x=\frac{-\left(-5\right)±3}{2}

Take the square root of 9.

x=\frac{5±3}{2}

The opposite of -5 is 5.

x=\frac{8}{2}

Now solve the equation x=\frac{5±3}{2} when ± is plus. Add 5 to 3.

x=4

Divide 8 by 2.

x=\frac{2}{2}

Now solve the equation x=\frac{5±3}{2} when ± is minus. Subtract 3 from 5.

x=1

Divide 2 by 2.

x=4 x=1

The equation is now solved.

x^{2}-5x+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.

x^{2}-5x+4-4=-4

Subtract 4 from both sides of the equation.

x^{2}-5x=-4

Subtracting 4 from itself leaves 0.

x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}=-2.5. Then add the square of -\frac{5}{2}=-2.5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}

Square -\frac{5}{2}=-2.5 by squaring both the numerator and the denominator of the fraction.

x^{2}-5x+\frac{25}{4}=\frac{9}{4}

Add -4 to \frac{25}{4}=6.25.

\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}

Factor x^{2}-5x+\frac{25}{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{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}

Take the square root of both sides of the equation.

x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}

Simplify.

x=4 x=1

Add \frac{5}{2}=2.5 to both sides of the equation.

x ^ 2 -5x +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.

r + s = 5 rs = 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{5}{2} - u s = \frac{5}{2} + u

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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{5}{2} - u) (\frac{5}{2} + u) = 4

To solve for unknown quantity u, substitute these in the product equation rs = 4

\frac{25}{4} - u^2 = 4

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 4-\frac{25}{4} = -\frac{9}{4}

Simplify the expression by subtracting \frac{25}{4} on both sides

u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =\frac{5}{2} - \frac{3}{2} = 1 s = \frac{5}{2} + \frac{3}{2} = 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

$x_{2}−4x−5=0$

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

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]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $