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Solve the quadratic equation 4 − 3x − 2x2 = 0 by completing the square.

Mathematics

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Step-by-step solution

14 steps

  1. Rearrange the equation into the standard form ax2+bx+c=0ax^2 + bx + c = 0.

    4−3x−2x2=04 - 3x - 2x^2 = 0 −2x2−3x+4=0-2x^2 - 3x + 4 = 0

    Multiply by −1-1 to make the leading coefficient positive:

    2x2+3x−4=02x^2 + 3x - 4 = 0

  2. Divide the entire equation by the coefficient of x2x^2, which is 2

    x2+32x−2=0x^2 + \frac{3}{2}x - 2 = 0

  3. Move the constant term to the right side of the equation

    x2+32x=2x^2 + \frac{3}{2}x = 2

  4. Complete the square on the left side.

    Take half of the coefficient of xx, square it, and add it to both sides.
    The coefficient of xx is 32\frac{3}{2}. Half of it is 12×32=34\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}.
    Squaring this gives (34)2=916\left(\frac{3}{4}\right)^2 = \frac{9}{16}.
    Add 916\frac{9}{16} to both sides:

    x2+32x+916=2+916x^2 + \frac{3}{2}x + \frac{9}{16} = 2 + \frac{9}{16}

  5. Factor the left side as a perfect square and simplify the right side

    (x+34)2=3216+916\left(x + \frac{3}{4}\right)^2 = \frac{32}{16} + \frac{9}{16} (x+34)2=4116\left(x + \frac{3}{4}\right)^2 = \frac{41}{16}

  6. Take the square root of both sides

    x+34=±4116x + \frac{3}{4} = \pm\sqrt{\frac{41}{16}} x+34=±414x + \frac{3}{4} = \pm\frac{\sqrt{41}}{4}

  7. Solve for xx

    x=−34±414x = -\frac{3}{4} \pm \frac{\sqrt{41}}{4} x=−3±414x = \frac{-3 \pm \sqrt{41}}{4}

    The final answer is x=−3±414\boxed{x = \frac{-3 \pm \sqrt{41}}{4}}.

  8. Substitute the given values of PP and QQ into the expression PQ\frac{P}{Q}

    PQ=4+22+2\frac{P}{Q} = \frac{4 + \sqrt{2}}{2 + \sqrt{2}}

  9. Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, which is 2−22 - \sqrt{2}

    PQ=4+22+2×2−22−2\frac{P}{Q} = \frac{4 + \sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}

  10. Expand the numerator and the denominator

    Numerator:

    (4+2)(2−2)=4(2)−42+22−(2)(2)(4 + \sqrt{2})(2 - \sqrt{2}) = 4(2) - 4\sqrt{2} + 2\sqrt{2} - (\sqrt{2})(\sqrt{2}) =8−42+22−2= 8 - 4\sqrt{2} + 2\sqrt{2} - 2 =6−22= 6 - 2\sqrt{2}

    Denominator (using the difference of squares formula (A+B)(A−B)=A2−B2(A+B)(A-B) = A^2 - B^2):

    (2+2)(2−2)=22−(2)2(2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - (\sqrt{2})^2 =4−2= 4 - 2 =2= 2

  11. Combine the simplified numerator and denominator

    PQ=6−222\frac{P}{Q} = \frac{6 - 2\sqrt{2}}{2}

  12. Simplify the expression by dividing both terms in the numerator by 2

    PQ=62−222\frac{P}{Q} = \frac{6}{2} - \frac{2\sqrt{2}}{2} PQ=3−2\frac{P}{Q} = 3 - \sqrt{2}

  13. Compare this result with the given form a+bca + b\sqrt{c}

    3−2=a+bc3 - \sqrt{2} = a + b\sqrt{c}

    By comparing the terms, we find:
    a=3a = 3
    b=−1b = -1
    c=2c = 2
    The final answer is a=3,b=−1,c=2\boxed{a = 3, b = -1, c = 2}.

  14. Identify the given temperature readings and their precision.

    P=22.5∘CP = 22.5^\circ C
    Q=19.4∘CQ = 19.4^\circ C

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