Answer :
To solve the equation \(6x^2 - 5x = 56\) using the zero product property, we first need to bring all terms to one side of the equation to set it to zero:
[tex]\[ 6x^2 - 5x - 56 = 0 \][/tex]
Next, we need to factorize the quadratic equation \(6x^2 - 5x - 56 = 0\). Factoring a quadratic in the form \(ax^2 + bx + c = 0\) involves finding two binomials \((mx + n)(px + q)\) such that their product gives back the quadratic expression.
However, this equation, \(6x^2 - 5x - 56\), does not factor easily by simple inspection, so we should typically use the quadratic formula instead:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \(a = 6\), \(b = -5\), and \(c = -56\). Plugging these values into the quadratic formula gives:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-56)}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 + 1344}}{12} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{1369}}{12} \][/tex]
[tex]\[ x = \frac{5 \pm 37}{12} \][/tex]
This results in two possible solutions:
1. \( x = \frac{5 + 37}{12} = \frac{42}{12} = 3.5 \)
2. \( x = \frac{5 - 37}{12} = \frac{-32}{12} = -\frac{8}{3} \)
So the solutions to the equation \(6x^2 - 5x = 56\) are:
[tex]\[ x = -\frac{8}{3} \quad \text{or} \quad x = 3.5 \][/tex]
Among the given options, the correct solution is:
[tex]\[ x = -\frac{8}{3} \quad \text{or} \quad x = \frac{7}{2} \][/tex]
Note that \(3.5\) can be written as a fraction \( \frac{7}{2}\).
Thus, the correct pair among the given choices is:
[tex]\[ x = -\frac{8}{3} \quad \text{or} \quad x = \frac{7}{2} \][/tex]
[tex]\[ 6x^2 - 5x - 56 = 0 \][/tex]
Next, we need to factorize the quadratic equation \(6x^2 - 5x - 56 = 0\). Factoring a quadratic in the form \(ax^2 + bx + c = 0\) involves finding two binomials \((mx + n)(px + q)\) such that their product gives back the quadratic expression.
However, this equation, \(6x^2 - 5x - 56\), does not factor easily by simple inspection, so we should typically use the quadratic formula instead:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \(a = 6\), \(b = -5\), and \(c = -56\). Plugging these values into the quadratic formula gives:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-56)}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 + 1344}}{12} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{1369}}{12} \][/tex]
[tex]\[ x = \frac{5 \pm 37}{12} \][/tex]
This results in two possible solutions:
1. \( x = \frac{5 + 37}{12} = \frac{42}{12} = 3.5 \)
2. \( x = \frac{5 - 37}{12} = \frac{-32}{12} = -\frac{8}{3} \)
So the solutions to the equation \(6x^2 - 5x = 56\) are:
[tex]\[ x = -\frac{8}{3} \quad \text{or} \quad x = 3.5 \][/tex]
Among the given options, the correct solution is:
[tex]\[ x = -\frac{8}{3} \quad \text{or} \quad x = \frac{7}{2} \][/tex]
Note that \(3.5\) can be written as a fraction \( \frac{7}{2}\).
Thus, the correct pair among the given choices is:
[tex]\[ x = -\frac{8}{3} \quad \text{or} \quad x = \frac{7}{2} \][/tex]
