Solve 2+14K+4K2=0

Mathematics

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Answer

−7±414\frac{-7 \pm \sqrt{41}}{4}

Step-by-step solution

6 steps

To solve the quadratic equation 4K2+14K+2=04K^2 + 14K + 2 = 0, we can use the quadratic formula K=−b±b2−4ac2aK = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

  1. Identify the coefficients aa, bb, and cc.

    The equation is in the form aK2+bK+c=0aK^2 + bK + c = 0.
    Here, a=4a=4, b=14b=14, and c=2c=2.

  2. Substitute the values into the quadratic formula.

    K=−(14)±(14)2−4(4)(2)2(4)K = \frac{-(14) \pm \sqrt{(14)^2 - 4(4)(2)}}{2(4)}

  3. Simplify the expression under the square root and the denominator.

    K=−14±196−328K = \frac{-14 \pm \sqrt{196 - 32}}{8} K=−14±1648K = \frac{-14 \pm \sqrt{164}}{8}

  4. Simplify the square root.

    We can factor out a perfect square from 164. 164=4×41164 = 4 \times 41.

    164=4×41=4×41=241\sqrt{164} = \sqrt{4 \times 41} = \sqrt{4} \times \sqrt{41} = 2\sqrt{41}

  5. Substitute the simplified square root back into the formula and simplify the expression.

    K=−14±2418K = \frac{-14 \pm 2\sqrt{41}}{8}

    Divide all terms by the common factor of 2.

    K=−7±414K = \frac{-7 \pm \sqrt{41}}{4}

  6. Write out the two solutions for KK.

    K1=−7+414K_1 = \frac{-7 + \sqrt{41}}{4} K2=−7−414K_2 = \frac{-7 - \sqrt{41}}{4}

    The solutions are:

    K=−7±414K = \boxed{\frac{-7 \pm \sqrt{41}}{4}}
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