Question

solve for x using the master product.
$3x^2 - 7x + 4 = 0$
master product:
answer attempt 1 out of 3
you must answer all questions above in order to submit.

Explanation:

Step1: Identify a, b, c

For quadratic \(ax^2 + bx + c = 0\), here \(a = 3\), \(b=-7\), \(c = 4\). Master product is \(a\times c=3\times4 = 12\). We need two numbers that multiply to 12 and add to \(b=-7\) (wait, no, for \(3x^2-7x + 4\), we split the middle term. Wait, actually, the master product method: for \(ax^2+bx + c\), find two numbers \(m\) and \(n\) such that \(m\times n=a\times c\) and \(m + n=b\). Wait, no, correction: for \(ax^2+bx + c\), when factoring, we want \(m\times n=a\times c\) and \(m + n=b\). Wait, here \(a = 3\), \(c = 4\), so \(a\times c=12\). We need two numbers that multiply to 12 and add to \(-7\)? Wait, no, the middle term is \(-7x\), so we need two numbers that multiply to \(3\times4 = 12\) and add to \(-7\)? Wait, no, actually, the correct approach: let's find two numbers that multiply to \(3\times4 = 12\) and add to \(-7\)? Wait, no, that can't be. Wait, maybe I messed up. Wait, the quadratic is \(3x^2-7x + 4\). Let's find two numbers that multiply to \(3\times4 = 12\) and add to \(-7\)? No, wait, the signs: since \(c\) is positive and \(b\) is negative, both numbers should be negative. Let's see: \(-3\) and \(-4\) multiply to 12 and add to \(-7\). Yes! So we split the middle term: \(3x^2-3x-4x + 4 = 0\). Then factor by grouping: \(3x(x - 1)-4(x - 1)=0\), so \((3x - 4)(x - 1)=0\). Then set each factor to zero: \(3x - 4 = 0\) or \(x - 1 = 0\).

Step2: Solve for x

For \(3x - 4 = 0\), \(3x=4\), so \(x=\frac{4}{3}\). For \(x - 1 = 0\), \(x = 1\).

Answer:

The solutions are \(x = 1\) and \(x=\frac{4}{3}\)