Using quadratic equation

Mathematics

ScanSolve AI Answer

Answer

3, -4

Step-by-step solution

5 steps

The equation you provided, 2x+x−12=02x+x-12=0, simplifies to 3x−12=03x-12=0. This is a linear equation, not a quadratic equation. The quadratic formula is used for equations of the form ax2+bx+c=0ax^2+bx+c=0, where a≠0a \neq 0.

However, based on our previous conversation for question 21, it's possible you meant to ask about the quadratic equation x2+x−12=0x^2+x-12=0. I will solve this equation using the quadratic formula.

Solving x2+x−12=0x^2+x-12=0 using the quadratic formula:

  1. Identify the coefficients.

    For a quadratic equation in the form ax2+bx+c=0ax^2+bx+c=0, we have:
    a=1a = 1
    b=1b = 1
    c=−12c = -12

  2. Apply the quadratic formula.

    The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.
    Substitute the values of aa, bb, and cc:

    x=−(1)±(1)2−4(1)(−12)2(1)x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-12)}}{2(1)}

  3. Simplify the expression under the square root.

    x=−1±1−(−48)2x = \frac{-1 \pm \sqrt{1 - (-48)}}{2} x=−1±1+482x = \frac{-1 \pm \sqrt{1 + 48}}{2} x=−1±492x = \frac{-1 \pm \sqrt{49}}{2}

  4. Calculate the square root.

    x=−1±72x = \frac{-1 \pm 7}{2}

  5. Find the two possible values for xx.

    For the positive root:

    x1=−1+72x_1 = \frac{-1 + 7}{2} x1=62x_1 = \frac{6}{2} x1=3x_1 = 3

    For the negative root:

    x2=−1−72x_2 = \frac{-1 - 7}{2} x2=−82x_2 = \frac{-8}{2} x2=−4x_2 = -4

    The possible values of xx are 3 and −4-4.

    The final answer is 3,−43, -4.

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