Question

determine the values of a, b, and c for the quadratic equation: ( x^2 - 6 = 3x )
options:
a=( x^2 ), b=-6, c=3x
a=1, b=-6, c=3
a=( x^2 ), b=3x, c=-6
a=1, b=-3, c=-6

Explanation:

Step1: Recall Quadratic Form

The standard quadratic equation is \(Ax^2 + Bx + C = 0\). We need to rearrange the given equation \(x^2 - 6 = 3x\) into this form.

Step2: Rearrange the Equation

Subtract \(3x\) from both sides: \(x^2 - 3x - 6 = 0\).

Step3: Identify Coefficients

Compare with \(Ax^2 + Bx + C = 0\). Here, \(A = 1\) (coefficient of \(x^2\)), \(B = -3\) (coefficient of \(x\)), and \(C = -6\)? Wait, no, wait the options: Wait, let's check the options again. Wait the equation is \(x^2 - 6 = 3x\), rearranged as \(x^2 - 3x - 6 = 0\)? Wait no, wait the options have \(A = 1\), \(B = -6\), \(C = 3\)? No, wait maybe I misread. Wait the correct rearrangement: \(x^2 - 3x - 6 = 0\)? Wait no, the given equation is \(x^2 - 6 = 3x\), so moving \(3x\) to left: \(x^2 - 3x - 6 = 0\). But looking at the options, one of them is \(A = 1\), \(B = -3\), \(C = -6\)? Wait the options: Let's check the visible options. The middle option (second one) is \(A = 1\), \(B = -6\), \(C = 3\)? No, wait the correct standard form is \(Ax^2 + Bx + C = 0\). So \(x^2 - 3x - 6 = 0\) would have \(A = 1\), \(B = -3\), \(C = -6\). But wait the options: Wait the last visible option is \(A = 1\), \(B = -3\), \(C = -6\)? Wait maybe the equation is \(x^2 - 6 = 3x\), so \(x^2 - 3x - 6 = 0\), so \(A = 1\), \(B = -3\), \(C = -6\). But looking at the options, the second option (with the crown) is \(A = 1\), \(B = -6\), \(C = 3\)? No, wait maybe I made a mistake. Wait the equation is \(x^2 - 6 = 3x\), so \(x^2 - 3x - 6 = 0\), so \(A = 1\), \(B = -3\), \(C = -6\). But the options: Let's check the options again. The first option: \(A = x^2\), \(B = -6\), \(C = 3x\) (wrong, \(A\) should be coefficient, not \(x^2\)). Second option: \(A = 1\), \(B = -6\), \(C = 3\) (wrong, because \(3x\) is on right, so moving to left gives \(-3x\)). Third option: \(A = x^2\), \(B = 3x\), \(C = -6\) (wrong, \(A\) should be coefficient). Fourth option: \(A = 1\), \(B = -3\), \(C = -6\) (correct, because \(x^2 - 3x - 6 = 0\)). Wait but the visible options: The last one is \(A = 1\), \(B = -3\), \(C = -6\)? So that's the correct one.

Answer:

The correct option is the last visible one (fourth, but in the image, the bottom option) with \(A = 1\), \(B = -3\), \(C = -6\) (or as per the options, the one with \(A = 1\), \(B = -3\), \(C = -6\)). But based on the standard form, the correct coefficients are \(A = 1\), \(B = -3\), \(C = -6\), so the option with \(A = 1\), \(B = -3\), \(C = -6\) (the last visible option in the image).