Quadratic Equations:x2 + 4 * x - 12 = 0Solve for $ x =\frac{ -b \pm \sqrt{b^2-4ac} }{2a} $Select all right answers for x

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BlackTom AI · Accepted Answer

We start by identifying the quadratic equation given: x^2 + 4x - 12 = 0. To find the roots, one standard method is factoring the quadratic.
Option a: x = 4. If we substitute x = 4 into the equation, we get 4^2 + 4·4 - 12 = 16 + 16 - 12 = 20, which is not zero. Therefore, x = 4 is not a solution.
Option b: x = -2. Substituting x = -2 yields (-2)^2 + 4·(-2) - 12 = 4 - 8 - 12 = -16, not zero. So x = -2 is not a solution.
Option c: x = 2. Substituting x = 2 gives 2^2 + 4·2 - 12 = 4 + 8 - 12 = 0, so x = 2 satisfies the equation. This is a valid root.
Option d: x = -6. Substituting x = -6 yields (-6)^2 + 4·(-6) - 12 = 36 - 24 - 12 = 0, so x = -6 also satisfies the equation. This is another valid root.
To connect with the quadratic formula: for x^2 + 4x - 12 = 0, a = 1, b = 4, c = -12. The discriminant b^2 - 4ac = 16 - 4(1)(-12) = 16 + 48 = 64, whose square root is 8. The roots are x = [-b ± sqrt(discriminant)]/(2a) = [-4 ± 8]/2. This yields x = (-4 + 8)/2 = 4/2 = 2 and x = (-4 - 8)/2 = -12/2 = -6. These results match the roots identified by factoring, confirming the correct values.
In summary, the two valid solutions from the list are x = 2 and x = -6, while x = 4 and x = -2 do not satisfy the equation.

HFe2_AMA FS24_A Practice 7

Quadratic Equations:x2 + 4 * x - 12 = 0Solve for \( x =\frac{ -b \pm \sqrt{b^2-4ac} }{2a} \)Select all right answers for x

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