Answer :
To solve the equation [tex]\( 3x^2 + 24x - 12 = 0 \)[/tex] by completing the square, follow these detailed steps:
1. Divide the entire equation by 3 to make the coefficient of [tex]\( x^2 \)[/tex] equal to 1:
[tex]\[ x^2 + 8x - 4 = 0 \][/tex]
2. Move the constant term to the other side of the equation:
[tex]\[ x^2 + 8x = 4 \][/tex]
3. Complete the square on the left side. To do this, add and subtract the square of half the coefficient of [tex]\( x \)[/tex], which is [tex]\( \left(\frac{8}{2}\right)^2 = 16 \)[/tex]:
[tex]\[ x^2 + 8x + 16 = 4 + 16 \][/tex]
Thus, the equation becomes:
[tex]\[ (x + 4)^2 = 20 \][/tex]
4. Take the square root of both sides:
[tex]\[ x + 4 = \pm\sqrt{20} \][/tex]
5. Solve for [tex]\( x \)[/tex] by isolating [tex]\( x \)[/tex]:
[tex]\[ x = -4 \pm \sqrt{20} \][/tex]
So, the solution to the equation [tex]\( 3x^2 + 24x - 12 = 0 \)[/tex] is:
[tex]\[ x = -4 \pm \sqrt{20} \][/tex]
- The constant term to be placed in Input Box 1 is -4.
- The number inside the radical to be placed in Input Box 2 is 20.
[tex]\[ \boxed{-4} \pm \sqrt{\boxed{20}} \][/tex]
1. Divide the entire equation by 3 to make the coefficient of [tex]\( x^2 \)[/tex] equal to 1:
[tex]\[ x^2 + 8x - 4 = 0 \][/tex]
2. Move the constant term to the other side of the equation:
[tex]\[ x^2 + 8x = 4 \][/tex]
3. Complete the square on the left side. To do this, add and subtract the square of half the coefficient of [tex]\( x \)[/tex], which is [tex]\( \left(\frac{8}{2}\right)^2 = 16 \)[/tex]:
[tex]\[ x^2 + 8x + 16 = 4 + 16 \][/tex]
Thus, the equation becomes:
[tex]\[ (x + 4)^2 = 20 \][/tex]
4. Take the square root of both sides:
[tex]\[ x + 4 = \pm\sqrt{20} \][/tex]
5. Solve for [tex]\( x \)[/tex] by isolating [tex]\( x \)[/tex]:
[tex]\[ x = -4 \pm \sqrt{20} \][/tex]
So, the solution to the equation [tex]\( 3x^2 + 24x - 12 = 0 \)[/tex] is:
[tex]\[ x = -4 \pm \sqrt{20} \][/tex]
- The constant term to be placed in Input Box 1 is -4.
- The number inside the radical to be placed in Input Box 2 is 20.
[tex]\[ \boxed{-4} \pm \sqrt{\boxed{20}} \][/tex]