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a+b=-2 ab=1
To solve the equation, factor a^{2}-2a+1 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.
a=-1 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(a-1\right)\left(a-1\right)
Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.
\left(a-1\right)^{2}
Rewrite as a binomial square.
a=1
To find equation solution, solve a-1=0.
a+b=-2 ab=1\times 1=1
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+1. To find a and b, set up a system to be solved.
a=-1 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(a^{2}-a\right)+\left(-a+1\right)
Rewrite a^{2}-2a+1 as \left(a^{2}-a\right)+\left(-a+1\right).
a\left(a-1\right)-\left(a-1\right)
Factor out a in the first and -1 in the second group.
\left(a-1\right)\left(a-1\right)
Factor out common term a-1 by using distributive property.
\left(a-1\right)^{2}
Rewrite as a binomial square.
a=1
To find equation solution, solve a-1=0.
a^{2}-2a+1=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.
a=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-2\right)±\sqrt{4-4}}{2}
Square -2.
a=\frac{-\left(-2\right)±\sqrt{0}}{2}
Add 4 to -4.
a=-\frac{-2}{2}
Take the square root of 0.
a=\frac{2}{2}
The opposite of -2 is 2.
a=1
Divide 2 by 2.
a^{2}-2a+1=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.
\left(a-1\right)^{2}=0
Factor a^{2}-2a+1. 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(a-1\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
a-1=0 a-1=0
Simplify.
a=1 a=1
Add 1 to both sides of the equation.
a=1
The equation is now solved. Solutions are the same.
x ^ 2 -2x +1 = 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.
r + s = 2 rs = 1
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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
1 - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-1 = 0
Simplify the expression by subtracting 1 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.