Multiple choice

Find the roots of the following quadratic equation by using the quadratic formula $(x^2 + 3x + 2)^2 - 8(x^2 + 3x) - 4 = 0$

  1. $0,1, 3, 4$,
  2. $0, 1, -3, 4$,
  3. $0, 1, -3, -4$

  4. None of these
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C Correct answer
Explanation

Let u = x^2 + 3x. The equation becomes (u+2)^2 - 8u - 4 = 0. u^2 + 4u + 4 - 8u - 4 = 0. u^2 - 4u = 0. u(u-4) = 0. So x^2 + 3x = 0 or x^2 + 3x = 4. Case 1: x(x+3) = 0 -> x=0, -3. Case 2: x^2 + 3x - 4 = 0 -> (x+4)(x-1) = 0 -> x=-4, 1. Roots are 0, 1, -3, -4.

AI explanation

Add 4 to both sides inside the parenthesis to form a perfect square, giving the equation the quantity x squared plus 3x plus 2 squared equals 12. Taking the square root of both sides means x squared plus 3x plus 2 equals 2 times the square root of 3 or negative 2 times the square root of 3. Solving both quadratic equations x squared plus 3x plus 2 minus 2 times the square root of 3 equals 0 and x squared plus 3x plus 2 plus 2 times the square root of 3 equals 0 using the quadratic formula yields the roots 0, 1, negative 3, and negative 4.

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