Solve for x
x=1
x=5
Graph
Share
Copied to clipboard
x^{2}-6x+5=0
Divide both sides by 3.
a+b=-6 ab=1\times 5=5
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.
a=-5 b=-1
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.
\left(x^{2}-5x\right)+\left(-x+5\right)
Rewrite x^{2}-6x+5 as \left(x^{2}-5x\right)+\left(-x+5\right).
x\left(x-5\right)-\left(x-5\right)
Factor out x in the first and -1 in the second group.
\left(x-5\right)\left(x-1\right)
Factor out common term x-5 by using distributive property.
x=5 x=1
To find equation solutions, solve x-5=0 and x-1=0.
3x^{2}-18x+15=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(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 15}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 15}}{2\times 3}
Square -18.
x=\frac{-\left(-18\right)±\sqrt{324-12\times 15}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-18\right)±\sqrt{324-180}}{2\times 3}
Multiply -12 times 15.
x=\frac{-\left(-18\right)±\sqrt{144}}{2\times 3}
Add 324 to -180.
x=\frac{-\left(-18\right)±12}{2\times 3}
Take the square root of 144.
x=\frac{18±12}{2\times 3}
The opposite of -18 is 18.
x=\frac{18±12}{6}
Multiply 2 times 3.
x=\frac{30}{6}
Now solve the equation x=\frac{18±12}{6} when ± is plus. Add 18 to 12.
x=5
Divide 30 by 6.
x=\frac{6}{6}
Now solve the equation x=\frac{18±12}{6} when ± is minus. Subtract 12 from 18.
x=1
Divide 6 by 6.
x=5 x=1
The equation is now solved.
3x^{2}-18x+15=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.
3x^{2}-18x+15-15=-15
Subtract 15 from both sides of the equation.
3x^{2}-18x=-15
Subtracting 15 from itself leaves 0.
\frac{3x^{2}-18x}{3}=-\frac{15}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{18}{3}\right)x=-\frac{15}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-6x=-\frac{15}{3}
Divide -18 by 3.
x^{2}-6x=-5
Divide -15 by 3.
x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-6x+9=-5+9
Square -3.
x^{2}-6x+9=4
Add -5 to 9.
\left(x-3\right)^{2}=4
Factor x^{2}-6x+9. 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-3\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x-3=2 x-3=-2
Simplify.
x=5 x=1
Add 3 to both sides of the equation.
x ^ 2 -6x +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 3
r + s = 6 rs = 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 = 3 - u s = 3 + u
Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>
(3 - u) (3 + u) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
9 - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-9 = -4
Simplify the expression by subtracting 9 on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =3 - 2 = 1 s = 3 + 2 = 5
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}