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Solve the quadratic equation using the quadratic formula 6x² − 4x + 9x − 6 = 0

Mathematics

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Answer

x=23,x=−32x = \frac{2}{3}, x = -\frac{3}{2}

Step-by-step solution

6 steps

To solve the quadratic equation 6x2−4x+9x−6=06x^2 - 4x + 9x - 6 = 0 using the quadratic formula, first simplify the equation.

  1. Combine like terms in the equation.

    6x2+(−4x+9x)−6=06x^2 + (-4x + 9x) - 6 = 0 6x2+5x−6=06x^2 + 5x - 6 = 0

  2. Identify the coefficients aa, bb, and cc from the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.

    For 6x2+5x−6=06x^2 + 5x - 6 = 0:
    a=6a = 6
    b=5b = 5
    c=−6c = -6

  3. Apply the quadratic formula.

    The quadratic formula is given by:

    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  4. Substitute the values of aa, bb, and cc into the formula.

    x=−(5)±(5)2−4(6)(−6)2(6)x = \frac{-(5) \pm \sqrt{(5)^2 - 4(6)(-6)}}{2(6)}

  5. Simplify the expression under the square root (the discriminant).

    x=−5±25−(−144)12x = \frac{-5 \pm \sqrt{25 - (-144)}}{12} x=−5±25+14412x = \frac{-5 \pm \sqrt{25 + 144}}{12} x=−5±16912x = \frac{-5 \pm \sqrt{169}}{12}

  6. Calculate the square root and find the two solutions for xx.

    Since 169=13\sqrt{169} = 13:

    x=−5±1312x = \frac{-5 \pm 13}{12}

    This gives two possible solutions:

    x1=−5+1312=812=23x_1 = \frac{-5 + 13}{12} = \frac{8}{12} = \frac{2}{3} x2=−5−1312=−1812=−32x_2 = \frac{-5 - 13}{12} = \frac{-18}{12} = -\frac{3}{2}

    The solutions are x=23x = \frac{2}{3} and x=−32x = -\frac{3}{2}.

    x=23,x=−32x = \frac{2}{3}, x = -\frac{3}{2}
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