Q. \(x^{2}+4x+3=0\)

Q: What are the roots of x^2+4x+3=0?

Factor: x^2+4x+3=(x+1)(x+3)=0. So x=-1 or x=-3.

Q: Can I solve it using the quadratic formula?

For a=1,b=4,c=3: x=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot3}}{2}=\frac{-4\pm\sqrt{4}}{2}=-2\pm1. Thus x=-1,-3.

Q: How do I factor x^2+4x+3?

Need numbers with sum 4 and product 3: 1 and 3. Since both are positive, x^2+4x+3=(x+1)(x+3).

Q: What is the discriminant and what does it mean here?

\Delta=b^2-4ac=16-12=4>0, so there are two distinct real solutions.

Q: Is there a way to complete the square?

x^2+4x+3=(x+2)^2-1=0. Then (x+2)^2=1, so x+2=\pm1, giving x=-1,-3.

Q: What is the sum and product of the solutions?

By Vieta’s formulas for x^2+4x+3=0: sum =-\frac{b}{a}=-4, product =\frac{c}{a}=3. Checks: (-1)+(-3)=-4, (-1)(-3)=3.

Answer

We solve \(x^2+4x+3=0\) by factoring:

\[
x^2+4x+3=(x+1)(x+3)=0
\]

So either \(x+1=0\) or \(x+3=0\).

\(x=-1\) or \(x=-3\).

Detailed Explanation

We want to solve the quadratic equation

\[
x^2 + 4x + 3 = 0
\]

Step 1: Factor the quadratic expression.

We look for two numbers that:

Multiply to give \(3\)
Add to give \(4\)

Those numbers are \(1\) and \(3\), because:

\[
1 \cdot 3 = 3 \quad \text{and} \quad 1 + 3 = 4
\]

So we factor the quadratic as:

\[
x^2 + 4x + 3 = (x+1)(x+3)
\]

Step 2: Set each factor equal to zero.

Using the zero product property, if

\[
(x+1)(x+3)=0
\]

then either

\[
x+1=0 \quad \text{or} \quad x+3=0
\]

Step 3: Solve each linear equation.

First equation:

\[
x+1=0
\]

Subtract \(1\) from both sides:

\[
x=-1
\]

Second equation:

\[
x+3=0
\]

Subtract \(3\) from both sides:

\[
x=-3
\]

Final Answer:

\[
x=-1 \quad \text{or} \quad x=-3
\]

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Algebra FAQ

What are the roots of \(x^2+4x+3=0\)?

Factor: \(x^2+4x+3=(x+1)(x+3)=0\). So \(x=-1\) or \(x=-3\).

Can I solve it using the quadratic formula?

For \(a=1,b=4,c=3\): \(x=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot3}}{2}=\frac{-4\pm\sqrt{4}}{2}=-2\pm1\). Thus \(x=-1,-3\).

How do I factor \(x^2+4x+3\)?

Need numbers with sum \(4\) and product \(3\): \(1\) and \(3\). Since both are positive, \(x^2+4x+3=(x+1)(x+3)\).

What is the discriminant and what does it mean here?

\(\Delta=b^2-4ac=16-12=4>0\), so there are two distinct real solutions.

Is there a way to complete the square?

\(x^2+4x+3=(x+2)^2-1=0\). Then \((x+2)^2=1\), so \(x+2=\pm1\), giving \(x=-1,-3\).

What is the sum and product of the solutions?

By Vieta’s formulas for \(x^2+4x+3=0\): sum \(=-\frac{b}{a}=-4\), product \(=\frac{c}{a}=3\). Checks: \((-1)+(-3)=-4\), \((-1)(-3)=3\).

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