Y 2 4x 3y 18
$\exponential{y}{ii} - 3 y - xviii $
\left(y-6\right)\left(y+3\right)
\left(y-6\right)\left(y+3\correct)
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a+b=-3 ab=ane\left(-18\right)=-eighteen
Factor the expression by grouping. First, the expression needs to be rewritten as y^{ii}+ay+past-18. To find a and b, set up a system to be solved.
1,-xviii 2,-9 iii,-6
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater accented value than the positive. List all such integer pairs that give product -18.
1-xviii=-17 2-9=-seven 3-6=-3
Summate the sum for each pair.
a=-6 b=3
The solution is the pair that gives sum -three.
\left(y^{2}-6y\right)+\left(3y-18\right)
Rewrite y^{2}-3y-18 as \left(y^{two}-6y\right)+\left(3y-18\right).
y\left(y-6\right)+iii\left(y-6\right)
Gene out y in the first and 3 in the 2nd group.
\left(y-6\correct)\left(y+3\right)
Factor out common term y-6 by using distributive property.
y^{2}-3y-18=0
Quadratic polynomial can be factored using the transformation ax^{two}+bx+c=a\left(x-x_{1}\right)\left(10-x_{ii}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{ii}+bx+c=0.
y=\frac{-\left(-iii\correct)±\sqrt{\left(-three\correct)^{2}-4\left(-18\right)}}{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 ii solutions, one when ± is addition and one when it is subtraction.
y=\frac{-\left(-3\right)±\sqrt{9-4\left(-eighteen\correct)}}{2}
Foursquare -3.
y=\frac{-\left(-3\right)±\sqrt{9+72}}{2}
Multiply -4 times -18.
y=\frac{-\left(-3\right)±\sqrt{81}}{two}
Add ix to 72.
y=\frac{-\left(-3\correct)±9}{2}
Have the square root of 81.
y=\frac{iii±9}{ii}
The opposite of -3 is iii.
y=\frac{12}{2}
Now solve the equation y=\frac{3±ix}{ii} when ± is plus. Add 3 to 9.
y=\frac{-vi}{two}
At present solve the equation y=\frac{3±9}{ii} when ± is minus. Subtract 9 from 3.
y^{2}-3y-18=\left(y-vi\correct)\left(y-\left(-iii\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{ane}\correct)\left(x-x_{2}\correct). Substitute 6 for x_{i} and -3 for x_{two}.
y^{2}-3y-xviii=\left(y-6\right)\left(y+3\right)
Simplify all the expressions of the class p-\left(-q\right) to p+q.
x ^ 2 -3x -18 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require gauge work. To utilise the direct factoring method, the equation must exist in the course x^2+Bx+C=0.
r + s = 3 rs = -18
Let r and s exist the factors for the quadratic equation such that 10^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the production of factors rs = C
r = \frac{three}{2} - u south = \frac{3}{2} + u
Two numbers r and s sum upwardly to 3 exactly when the boilerplate of the two numbers is \frac{1}{2}*three = \frac{3}{2}. You tin can as well see that the midpoint of r and southward corresponds to the axis of symmetry of the parabola represented past the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the heart 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' style='width: 100%;max-width: 700px' /></div>
(\frac{3}{2} - u) (\frac{three}{2} + u) = -18
To solve for unknown quantity u, substitute these in the product equation rs = -18
\frac{9}{iv} - u^2 = -18
Simplify by expanding (a -b) (a + b) = a^2 – b^ii
-u^2 = -18-\frac{9}{4} = -\frac{81}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^two = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{two}
Simplify the expression by multiplying -1 on both sides and accept the square root to obtain the value of unknown variable u
r =\frac{iii}{2} - \frac{9}{2} = -3 s = \frac{3}{2} + \frac{ix}{two} = half-dozen
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Y 2 4x 3y 18,
Source: https://mathsolver.microsoft.com/en/solve-problem/y%20%5E%20%7B%202%20%7D%20-%203%20y%20-%2018
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