Give me a quadratic equation question

Mathematics

ScanSolve AI Answer

Answer

x=12,x=−3x = \frac{1}{2}, x = -3

Step-by-step solution

5 steps

Solve for xx: 2x2+5x−3=02x^2 + 5x - 3 = 0

  1. Use the quadratic formula, which is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    In this equation, a=2a=2, b=5b=5, and c=−3c=-3.

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

    x=−5±52−4(2)(−3)2(2)x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}

  3. Simplify the expression under the square root.

    x=−5±25−(−24)4x = \frac{-5 \pm \sqrt{25 - (-24)}}{4} x=−5±25+244x = \frac{-5 \pm \sqrt{25 + 24}}{4} x=−5±494x = \frac{-5 \pm \sqrt{49}}{4}

  4. Calculate the square root of 49.

    x=−5±74x = \frac{-5 \pm 7}{4}

  5. Find the two possible values for xx.

    For the '+' sign:

    x1=−5+74=24=12x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}

    For the '-' sign:

    x2=−5−74=−124=−3x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3

    The solutions are x=12x = \frac{1}{2} and x=−3x = -3.

    x=12,x=−3x = \frac{1}{2}, x = -3
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