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Solve the following quadratic equations using the quadratic formula: 1) x2 + 3x + 1 = 0 2) x2 − 7x − 2 = 0 3) 2m2 + 10m − 5 = 0

Mathematics

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We will solve each quadratic equation using the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

1) x2+3x+1=0x^2 + 3x + 1 = 0

  1. Identify the coefficients.

    Here, a=1a=1, b=3b=3, c=1c=1.

  2. Substitute the values into the quadratic formula.

    x=−3±32−4(1)(1)2(1)x = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2(1)} x=−3±9−42x = \frac{-3 \pm \sqrt{9 - 4}}{2} x=−3±52x = \frac{-3 \pm \sqrt{5}}{2}
  3. Write the two solutions.

    x1=−3+52x_1 = \frac{-3 + \sqrt{5}}{2} x2=−3−52x_2 = \frac{-3 - \sqrt{5}}{2}

    The solutions are x=−3±52\boxed{x = \frac{-3 \pm \sqrt{5}}{2}}.

    2) x2−7x−2=0x^2 - 7x - 2 = 0

  4. Identify the coefficients.

    Here, a=1a=1, b=−7b=-7, c=−2c=-2.

  5. Substitute the values into the quadratic formula.

    x=−(−7)±(−7)2−4(1)(−2)2(1)x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-2)}}{2(1)} x=7±49+82x = \frac{7 \pm \sqrt{49 + 8}}{2} x=7±572x = \frac{7 \pm \sqrt{57}}{2}
  6. Write the two solutions.

    x1=7+572x_1 = \frac{7 + \sqrt{57}}{2} x2=7−572x_2 = \frac{7 - \sqrt{57}}{2}

    The solutions are x=7±572\boxed{x = \frac{7 \pm \sqrt{57}}{2}}.

    3) 2m2+10m−5=02m^2 + 10m - 5 = 0

  7. Identify the coefficients.

    Here, a=2a=2, b=10b=10, c=−5c=-5.

  8. Substitute the values into the quadratic formula.

    m=−10±102−4(2)(−5)2(2)m = \frac{-10 \pm \sqrt{10^2 - 4(2)(-5)}}{2(2)} m=−10±100+404m = \frac{-10 \pm \sqrt{100 + 40}}{4} m=−10±1404m = \frac{-10 \pm \sqrt{140}}{4}
  9. Simplify the radical 140\sqrt{140}.

    140=4×35=235\sqrt{140} = \sqrt{4 \times 35} = 2\sqrt{35}
  10. Substitute back and simplify the expression.

    m=−10±2354m = \frac{-10 \pm 2\sqrt{35}}{4} m=2(−5±35)4m = \frac{2(-5 \pm \sqrt{35})}{4} m=−5±352m = \frac{-5 \pm \sqrt{35}}{2}

    The solutions are m=−5±352\boxed{m = \frac{-5 \pm \sqrt{35}}{2}}.

    4) 5q2−q−3=05q^2 - q - 3 = 0

  11. Identify the coefficients.

    Here, a=5a=5, b=−1b=-1, c=−3c=-3.

  12. Substitute the values into the quadratic formula.

    q=−(−1)±(−1)2−4(5)(−3)2(5)q = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(5)(-3)}}{2(5)} q=1±1+6010q = \frac{1 \pm \sqrt{1 + 60}}{10} q=1±6110q = \frac{1 \pm \sqrt{61}}{10}
  13. Write the two solutions.

    q1=1+6110q_1 = \frac{1 + \sqrt{61}}{10} q2=1−6110q_2 = \frac{1 - \sqrt{61}}{10}

    The solutions are q=1±6110\boxed{q = \frac{1 \pm \sqrt{61}}{10}}.

    5) 10x2−4x=1610x^2 - 4x = 16

  14. Rewrite the equation in standard form ax2+bx+c=0ax^2 + bx + c = 0.

    10x2−4x−16=010x^2 - 4x - 16 = 0

    Divide by 2 to simplify:

    5x2−2x−8=05x^2 - 2x - 8 = 0

    Now, identify the coefficients: a=5a=5, b=−2b=-2, c=−8c=-8.

  15. Substitute the values into the quadratic formula.

    x=−(−2)±(−2)2−4(5)(−8)2(5)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(-8)}}{2(5)} x=2±4+16010x = \frac{2 \pm \sqrt{4 + 160}}{10} x=2±16410x = \frac{2 \pm \sqrt{164}}{10}
  16. Simplify the radical 164\sqrt{164}.

    164=4×41=241\sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41}
  17. Substitute back and simplify the expression.

    x=2±24110x = \frac{2 \pm 2\sqrt{41}}{10} x=2(1±41)10x = \frac{2(1 \pm \sqrt{41})}{10} x=1±415x = \frac{1 \pm \sqrt{41}}{5}

    The solutions are x=1±415\boxed{x = \frac{1 \pm \sqrt{41}}{5}}.

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