Solve Quadratic Equation: Finding Solutions to x²+4x+3=0

Let's solve the given equation:

$x^2+4x+3=0$

We identify that the coefficient of the quadratic term is 1, therefore we can (try to) solve it using the quick trinomial method,

We'll look for a pair of numbers whose product is the free term in the left side expression, and whose sum is the coefficient of the first power term meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=3\\ m+n=4$

From the first requirement above, meaning - from the multiplication, we can conclude according to the rules of sign multiplication that both numbers have the same sign, and now we'll remember that the possible factor pairs of the (prime) number 3 are 3 and 1, fulfilling the second requirement mentioned, together with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=3\\ n=1 \end{cases}$

and therefore we'll factor the expression on the left side of the equation to:

$x^2+4x+3=0\\ \downarrow\\ (x+3)(x+1)=0$

where we used the pair of numbers we found earlier in this factorization,

We'll continue and consider the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x+3=0\\ \boxed{x=-3}$

or:

$x+1=0\\ \boxed{x=-1}$

Therefore the given equation has two real solutions.

Therefore the correct answer is answer B.

Important note:

Note that we could answer the question asked in the problem and find the number of real solutions to the equation (and if there are any real solutions to this equation at all), without solving it,

For this we'll recall the roots formula:

The rule states that for a quadratic equation in the general form:

$ax^2+bx+c =0$there are two solutions (or fewer) which we find using the formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

We can now denote the expression under the root in the general equation solutions as:

$\Delta$ (The Greek letter delta), meaning, mathematically:

$x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac$

We can now recall that a real root can only be calculated for a positive number or zero (since you cannot get a negative number by multiplying a real number by itself), and therefore an equation has two real solutions (or one real solution) only if:

$\Delta\geq0$

Later we'll note that if: $\Delta=0$ then the only solution to the equation is:

$x=-\frac{b}{2a}$

Therefore, we can summarize what was explained into a rule:

In a quadratic equation:

$ax^2+bx+c =0$

where we extract the coefficients and calculate delta:

$\Delta=b^2-4ac$

If:

a.$\Delta<0$:

There is no (real) solution to the equation.

b.$\Delta=0$:

There is one (real) solution to the equation.

c.$\Delta=0$:

There are two (real) solutions to the equation.

Now let's return to the given equation and extract from it the coefficients:

$x^2+4x+3=0 \longrightarrow\begin{cases}a=1\\b=4\\c=3\end{cases}$We'll continue and calculate $\Delta$:

$\Delta=b^2-4ac\longrightarrow \Delta=4^2-4\cdot1\cdot3=4\\ \downarrow\\ \boxed{\Delta=4}$meaning we got that:

$\boxed{ \Delta>0}$Therefore the given equation has two real solutions.