X 2 6x 5 0
$\exponential{(x)}{ii} + half-dozen x + five = 0 $
x=-v
ten=-ane
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a+b=6 ab=5
To solve the equation, factor x^{2}+6x+5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(ten+b\right). To detect a and b, ready a arrangement to exist solved.
a=ane b=5
Since ab is positive, a and b have the aforementioned sign. Since a+b is positive, a and b are both positive. The just such pair is the system solution.
\left(10+i\right)\left(x+5\correct)
Rewrite factored expression \left(x+a\correct)\left(ten+b\right) using the obtained values.
x=-1 x=-5
To find equation solutions, solve x+one=0 and x+five=0.
a+b=half-dozen ab=1\times 5=5
To solve the equation, gene the left hand side by grouping. First, left hand side needs to be rewritten as ten^{two}+ax+bx+v. To find a and b, set up a system to exist solved.
a=1 b=5
Since ab is positive, a and b take the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
\left(x^{2}+x\right)+\left(5x+five\right)
Rewrite x^{2}+6x+5 equally \left(x^{2}+ten\right)+\left(5x+5\correct).
x\left(x+one\right)+v\left(x+1\correct)
Factor out x in the first and 5 in the 2d grouping.
\left(ten+1\right)\left(x+5\right)
Gene out common term x+one past using distributive property.
x=-one x=-5
To find equation solutions, solve x+1=0 and x+5=0.
x^{2}+6x+5=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{two}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when information technology is subtraction.
10=\frac{-6±\sqrt{6^{2}-4\times 5}}{2}
This equation is in standard course: ax^{2}+bx+c=0. Substitute ane for a, 6 for b, and five for c in the quadratic formula, \frac{-b±\sqrt{b^{two}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times v}}{2}
Square half dozen.
x=\frac{-6±\sqrt{36-20}}{ii}
Multiply -4 times 5.
x=\frac{-6±\sqrt{xvi}}{two}
Add together 36 to -20.
ten=\frac{-6±4}{2}
Take the square root of 16.
ten=\frac{-2}{2}
Now solve the equation x=\frac{-vi±4}{2} when ± is plus. Add together -half-dozen to 4.
x=\frac{-x}{2}
Now solve the equation x=\frac{-vi±four}{2} when ± is minus. Subtract 4 from -half-dozen.
10=-i ten=-five
The equation is now solved.
10^{2}+6x+v=0
Quadratic equations such as this 1 can exist solved by completing the square. In order to complete the square, the equation must offset be in the form x^{2}+bx=c.
x^{2}+6x+5-v=-5
Subtract 5 from both sides of the equation.
ten^{two}+6x=-v
Subtracting v from itself leaves 0.
x^{2}+6x+3^{ii}=-five+three^{2}
Divide 6, the coefficient of the x term, by 2 to get iii. Then add the square of 3 to both sides of the equation. This pace makes the left hand side of the equation a perfect square.
10^{ii}+6x+9=-5+nine
Square 3.
ten^{2}+6x+9=four
Add -5 to 9.
\left(10+iii\right)^{2}=4
Factor x^{2}+6x+ix. In full general, when x^{2}+bx+c is a perfect square, it can e'er exist factored every bit \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+iii\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
10=-ane x=-5
Subtract 3 from both sides of the equation.
x ^ 2 +6x +5 = 0
Quadratic equations such as this one can be solved by a new straight factoring method that does not require guess work. To employ the directly factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = -6 rs = 5
Permit r and south be the factors for the quadratic equation such that ten^2+Bx+C=(10−r)(ten−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 southward sum upward to -six exactly when the average of the ii numbers is \frac{1}{two}*-6 = -3. You can also see that the midpoint of r and southward corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and due south 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 production equation rs = v
9 - u^2 = 5
Simplify past expanding (a -b) (a + b) = a^two – b^2
-u^2 = 5-9 = -four
Simplify the expression past subtracting nine on both sides
u^2 = 4 u = \pm\sqrt{iv} = \pm 2
Simplify the expression past multiplying -1 on both sides and take the foursquare root to obtain the value of unknown variable u
r =-3 - ii = -v s = -iii + 2 = -1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and south.
X 2 6x 5 0,
Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20%2B6x%2B5%20%3D%20%200
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