Question

1. $3x^2 + 7x + 4 = 0$

Explanation:

Step1: Factor the quadratic equation

We need to factor \(3x^2 + 7x + 4\). We look for two numbers that multiply to \(3\times4 = 12\) and add up to \(7\). The numbers are \(3\) and \(4\).
So we rewrite the middle term: \(3x^2+3x + 4x + 4 = 0\)
Group the terms: \((3x^2 + 3x)+(4x + 4)=0\)
Factor out the common factors: \(3x(x + 1)+4(x + 1)=0\)
Then we get \((3x + 4)(x + 1)=0\)

Step2: Solve for x

Set each factor equal to zero:
\(3x+4 = 0\) or \(x + 1 = 0\)
For \(3x+4 = 0\), subtract \(4\) from both sides: \(3x=-4\), then divide by \(3\): \(x=-\frac{4}{3}\)
For \(x + 1 = 0\), subtract \(1\) from both sides: \(x=-1\)

Answer:

\(x=-1\) or \(x =-\frac{4}{3}\)