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Solve quadratic equations i - ii using completing square method and iii - iv using factorization methods.

Mathematics

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19 steps

let's solve these quadratic equations using the specified methods.

Part i· 6 steps

Solve ax2+bx+c=0ax^2 + bx + c = 0 using the completing the square method.

  1. Divide the entire equation by aa (assuming a≠0a \neq 0).

    x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0

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

    x2+bax=−cax^2 + \frac{b}{a}x = -\frac{c}{a}

  3. Complete the square on the left side.

    To do this, add (12⋅ba)2=(b2a)2=b24a2\left(\frac{1}{2} \cdot \frac{b}{a}\right)^2 = \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} to both sides of the equation.

    x2+bax+b24a2=−ca+b24a2x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}

  4. Factor the left side as a perfect square and combine the terms on the right side.

    (x+b2a)2=b2−4ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

  5. Take the square root of both sides.

    x+b2a=±b2−4ac4a2x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} x+b2a=±b2−4ac2ax + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}

  6. Isolate xx to find the solutions.

    x=−b2a±b2−4ac2ax = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} x=−b±b2−4ac2ax = \boxed{\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}

Part ii· 5 steps

Solve 4.6x2+3.5x−1.75=04.6x^2 + 3.5x - 1.75 = 0 using the completing the square method.

  1. Divide the entire equation by 4.6.

    x2+3.54.6x−1.754.6=0x^2 + \frac{3.5}{4.6}x - \frac{1.75}{4.6} = 0

    To work with exact values, convert decimals to fractions: 4.6=2354.6 = \frac{23}{5}, 3.5=723.5 = \frac{7}{2}, 1.75=741.75 = \frac{7}{4}.
    The equation becomes:

    235x2+72x−74=0\frac{23}{5}x^2 + \frac{7}{2}x - \frac{7}{4} = 0

    Multiply by 20 to clear denominators:

    92x2+70x−35=092x^2 + 70x - 35 = 0

    Now divide by 92:

    x2+7092x−3592=0x^2 + \frac{70}{92}x - \frac{35}{92} = 0 x2+3546x−3592=0x^2 + \frac{35}{46}x - \frac{35}{92} = 0

  2. Move the constant term to the right side.

    x2+3546x=3592x^2 + \frac{35}{46}x = \frac{35}{92}

  3. Complete the square.

    Add (12⋅3546)2=(3592)2=12258464\left(\frac{1}{2} \cdot \frac{35}{46}\right)^2 = \left(\frac{35}{92}\right)^2 = \frac{1225}{8464} to both sides.

    x2+3546x+12258464=3592+12258464x^2 + \frac{35}{46}x + \frac{1225}{8464} = \frac{35}{92} + \frac{1225}{8464} (x+3592)2=35×9292×92+12258464\left(x + \frac{35}{92}\right)^2 = \frac{35 \times 92}{92 \times 92} + \frac{1225}{8464} (x+3592)2=32208464+12258464\left(x + \frac{35}{92}\right)^2 = \frac{3220}{8464} + \frac{1225}{8464} (x+3592)2=44458464\left(x + \frac{35}{92}\right)^2 = \frac{4445}{8464}

  4. Take the square root of both sides.

    x+3592=±44458464x + \frac{35}{92} = \pm\sqrt{\frac{4445}{8464}} x+3592=±444592x + \frac{35}{92} = \pm\frac{\sqrt{4445}}{92}

  5. Isolate xx.

    x=−3592±444592x = -\frac{35}{92} \pm \frac{\sqrt{4445}}{92} x=−35±444592x = \frac{-35 \pm \sqrt{4445}}{92}

    Using 4445≈66.6708\sqrt{4445} \approx 66.6708:

    x1=−35+66.670892=31.670892≈0.3442x_1 = \frac{-35 + 66.6708}{92} = \frac{31.6708}{92} \approx 0.3442 x2=−35−66.670892=−101.670892≈−1.1051x_2 = \frac{-35 - 66.6708}{92} = \frac{-101.6708}{92} \approx -1.1051 x=−35±444592 or approximately x1≈0.34,x2≈−1.11x = \boxed{\frac{-35 \pm \sqrt{4445}}{92} \text{ or approximately } x_1 \approx 0.34, x_2 \approx -1.11}

Part iii· 5 steps

Solve 15x2+2x−8=015x^2 + 2x - 8 = 0 using factorization.

  1. Find two numbers that multiply to ac=15×(−8)=−120ac = 15 \times (-8) = -120 and add up to b=2b = 2.

    The numbers are 12 and −10-10. (12×−10=−12012 \times -10 = -120 and 12+(−10)=212 + (-10) = 2).

  2. Rewrite the middle term ($2x$) using these two numbers.

    15x2+12x−10x−8=015x^2 + 12x - 10x - 8 = 0

  3. Factor by grouping the terms.

    (15x2+12x)−(10x+8)=0(15x^2 + 12x) - (10x + 8) = 0 3x(5x+4)−2(5x+4)=03x(5x + 4) - 2(5x + 4) = 0

  4. Factor out the common binomial (5x+4)(5x+4).

    (3x−2)(5x+4)=0(3x - 2)(5x + 4) = 0

  5. Set each factor to zero and solve for xx.

    3x−2=0⇒3x=2⇒x=233x - 2 = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3} 5x+4=0⇒5x=−4⇒x=−455x + 4 = 0 \Rightarrow 5x = -4 \Rightarrow x = -\frac{4}{5} x=23,−45x = \boxed{\frac{2}{3}, -\frac{4}{5}}

Part iv· 3 steps

Solve 4x2−25=04x^2 - 25 = 0 using factorization.

  1. Recognize the equation as a difference of squares, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

    Here, a2=4x2a^2 = 4x^2, so a=4x2=2xa = \sqrt{4x^2} = 2x.
    And b2=25b^2 = 25, so b=25=5b = \sqrt{25} = 5.

  2. Apply the difference of squares formula.

    (2x−5)(2x+5)=0(2x - 5)(2x + 5) = 0

  3. Set each factor to zero and solve for xx.

    2x−5=0⇒2x=5⇒x=522x - 5 = 0 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2} 2x+5=0⇒2x=−5⇒x=−522x + 5 = 0 \Rightarrow 2x = -5 \Rightarrow x = -\frac{5}{2} x=52,−52x = \boxed{\frac{5}{2}, -\frac{5}{2}}
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