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a+b=-7 ab=5\times 2=10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5u^{2}+au+bu+2. To find a and b, set up a system to be solved.
-1,-10 -2,-5
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 10.
-1-10=-11 -2-5=-7
Calculate the sum for each pair.
a=-5 b=-2
The solution is the pair that gives sum -7.
\left(5u^{2}-5u\right)+\left(-2u+2\right)
Rewrite 5u^{2}-7u+2 as \left(5u^{2}-5u\right)+\left(-2u+2\right).
5u\left(u-1\right)-2\left(u-1\right)
Factor out 5u in the first and -2 in the second group.
\left(u-1\right)\left(5u-2\right)
Factor out common term u-1 by using distributive property.
u=1 u=\frac{2}{5}
To find equation solutions, solve u-1=0 and 5u-2=0.
5u^{2}-7u+2=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.
u=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\times 2}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
u=\frac{-\left(-7\right)±\sqrt{49-4\times 5\times 2}}{2\times 5}
Square -7.
u=\frac{-\left(-7\right)±\sqrt{49-20\times 2}}{2\times 5}
Multiply -4 times 5.
u=\frac{-\left(-7\right)±\sqrt{49-40}}{2\times 5}
Multiply -20 times 2.
u=\frac{-\left(-7\right)±\sqrt{9}}{2\times 5}
Add 49 to -40.
u=\frac{-\left(-7\right)±3}{2\times 5}
Take the square root of 9.
u=\frac{7±3}{2\times 5}
The opposite of -7 is 7.
u=\frac{7±3}{10}
Multiply 2 times 5.
u=\frac{10}{10}
Now solve the equation u=\frac{7±3}{10} when ± is plus. Add 7 to 3.
u=1
Divide 10 by 10.
u=\frac{4}{10}
Now solve the equation u=\frac{7±3}{10} when ± is minus. Subtract 3 from 7.
u=\frac{2}{5}
Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.
u=1 u=\frac{2}{5}
The equation is now solved.
5u^{2}-7u+2=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.
5u^{2}-7u+2-2=-2
Subtract 2 from both sides of the equation.
5u^{2}-7u=-2
Subtracting 2 from itself leaves 0.
\frac{5u^{2}-7u}{5}=-\frac{2}{5}
Divide both sides by 5.
u^{2}-\frac{7}{5}u=-\frac{2}{5}
Dividing by 5 undoes the multiplication by 5.
u^{2}-\frac{7}{5}u+\left(-\frac{7}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{7}{10}\right)^{2}
Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
u^{2}-\frac{7}{5}u+\frac{49}{100}=-\frac{2}{5}+\frac{49}{100}
Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.
u^{2}-\frac{7}{5}u+\frac{49}{100}=\frac{9}{100}
Add -\frac{2}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(u-\frac{7}{10}\right)^{2}=\frac{9}{100}
Factor u^{2}-\frac{7}{5}u+\frac{49}{100}. 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(u-\frac{7}{10}\right)^{2}}=\sqrt{\frac{9}{100}}
Take the square root of both sides of the equation.
u-\frac{7}{10}=\frac{3}{10} u-\frac{7}{10}=-\frac{3}{10}
Simplify.
u=1 u=\frac{2}{5}
Add \frac{7}{10} to both sides of the equation.
x ^ 2 -\frac{7}{5}x +\frac{2}{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 5
r + s = \frac{7}{5} rs = \frac{2}{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 = \frac{7}{10} - u s = \frac{7}{10} + u
Two numbers r and s sum up to \frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{5} = \frac{7}{10}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{7}{10} - u) (\frac{7}{10} + u) = \frac{2}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}
\frac{49}{100} - u^2 = \frac{2}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{5}-\frac{49}{100} = -\frac{9}{100}
Simplify the expression by subtracting \frac{49}{100} on both sides
u^2 = \frac{9}{100} u = \pm\sqrt{\frac{9}{100}} = \pm \frac{3}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{10} - \frac{3}{10} = 0.400 s = \frac{7}{10} + \frac{3}{10} = 1.000
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.