6 X 9 X 5
$\exponential{(x)}{two} - 6 10 + 9 $
\left(x-iii\right)^{2}
\left(x-three\correct)^{2}
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a+b=-6 ab=1\times ix=9
Factor the expression by grouping. Start, the expression needs to be rewritten every bit ten^{2}+ax+bx+nine. To observe a and b, set up a system to be solved.
-1,-9 -3,-3
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 9.
-one-9=-10 -3-3=-6
Summate the sum for each pair.
a=-3 b=-3
The solution is the pair that gives sum -6.
\left(x^{two}-3x\right)+\left(-3x+nine\correct)
Rewrite x^{2}-6x+9 as \left(x^{2}-3x\right)+\left(-3x+9\correct).
ten\left(x-three\right)-3\left(10-three\right)
Factor out ten in the first and -3 in the second grouping.
\left(x-3\correct)\left(x-three\correct)
Factor out common term x-three by using distributive holding.
\left(x-3\correct)^{2}
Rewrite as a binomial square.
factor(ten^{2}-6x+9)
This trinomial has the course of a trinomial square, perhaps multiplied past a common factor. Trinomial squares can be factored past finding the square roots of the leading and trailing terms.
\sqrt{9}=iii
Notice the foursquare root of the trailing term, 9.
\left(10-3\right)^{2}
The trinomial square is the square of the binomial that is the sum or departure of the square roots of the leading and abaft terms, with the sign determined by the sign of the middle term of the trinomial square.
ten^{2}-6x+ix=0
Quadratic polynomial tin can exist factored using the transformation ax^{two}+bx+c=a\left(x-x_{1}\right)\left(10-x_{2}\right), where x_{1} and x_{ii} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-6\correct)±\sqrt{\left(-half-dozen\right)^{two}-4\times 9}}{2}
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, 1 when ± is addition and one when it is subtraction.
ten=\frac{-\left(-6\correct)±\sqrt{36-four\times 9}}{2}
Square -half-dozen.
x=\frac{-\left(-6\right)±\sqrt{36-36}}{two}
Multiply -4 times 9.
x=\frac{-\left(-half-dozen\right)±\sqrt{0}}{2}
Add 36 to -36.
ten=\frac{-\left(-vi\correct)±0}{2}
Accept the foursquare root of 0.
x=\frac{6±0}{2}
The opposite of -half-dozen is 6.
ten^{2}-6x+ix=\left(x-3\right)\left(ten-iii\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\correct)\left(x-x_{2}\right). Substitute 3 for x_{1} and iii for x_{2}.
10 ^ 2 -6x +ix = 0
Quadratic equations such as this one can be solved past a new straight factoring method that does not require guess piece of work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + southward = 6 rs = nine
Let r and s exist 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 southward = 3 + u
Two numbers r and s sum up to 6 exactly when the average of the ii numbers is \frac{one}{2}*6 = three. Yous can as well encounter 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 past 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' way='width: 100%;max-width: 700px' /></div>
(three - u) (3 + u) = 9
To solve for unknown quantity u, substitute these in the product equation rs = 9
9 - u^2 = 9
Simplify past expanding (a -b) (a + b) = a^2 – b^2
-u^two = 9-9 = 0
Simplify the expression by subtracting 9 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and accept the square root to obtain the value of unknown variable u
r = southward = 3
The factors r and due south are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
6 X 9 X 5,
Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20-6x%2B9
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