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a+b=4 ab=3
To solve the equation, factor x^{2}+4x+3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
a=1 b=3
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
\left(x+1\right)\left(x+3\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=-1 x=-3
To find equation solutions, solve x+1=0 and x+3=0.
a+b=4 ab=1\times 3=3
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.
a=1 b=3
Since ab is positive, a and b have 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(3x+3\right)
Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).
x\left(x+1\right)+3\left(x+1\right)
Factor out x in the first and 3 in the second group.
\left(x+1\right)\left(x+3\right)
Factor out common term x+1 by using distributive property.
x=-1 x=-3
To find equation solutions, solve x+1=0 and x+3=0.
x^{2}+4x+3=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.
x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\times 3}}{2}
Square 4.
x=\frac{-4±\sqrt{16-12}}{2}
Multiply -4 times 3.
x=\frac{-4±\sqrt{4}}{2}
Add 16 to -12.
x=\frac{-4±2}{2}
Take the square root of 4.
x=-\frac{2}{2}
Now solve the equation x=\frac{-4±2}{2} when ± is plus. Add -4 to 2.
x=-1
Divide -2 by 2.
x=-\frac{6}{2}
Now solve the equation x=\frac{-4±2}{2} when ± is minus. Subtract 2 from -4.
x=-3
Divide -6 by 2.
x=-1 x=-3
The equation is now solved.
x^{2}+4x+3=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.
x^{2}+4x+3-3=-3
Subtract 3 from both sides of the equation.
x^{2}+4x=-3
Subtracting 3 from itself leaves 0.
x^{2}+4x+2^{2}=-3+2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+4x+4=-3+4
Square 2.
x^{2}+4x+4=1
Add -3 to 4.
\left(x+2\right)^{2}=1
Factor x^{2}+4x+4. 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(x+2\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x+2=1 x+2=-1
Simplify.
x=-1 x=-3
Subtract 2 from both sides of the equation.