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Rafimarujigo
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Find the solution of each of the following quadratic by completing the square
[tex]2 \times {}^{2} - 16 \times = - 14[/tex]
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KAntoineDoixgo

✏️QUADRATIC

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[tex] \large \bold{\blue{PROBLEM:}} [/tex] Find the solution of each of the following quadratic by completing the square

  • [tex] 2x^2 - 16x = - 14 [/tex]

[tex] \large \bold{\blue{SOLUTION:}} [/tex] Divide both sides of the equation by the a to isolate the squared variable and to simplify (it only depends if the constant and the coefficient is divisible by a)

  • [tex] \frac{2x^2 - 16x}{2} = \frac{-14}{2} \\ [/tex]

  • [tex] x^2 - 8x = -7 [/tex]

» The equation has been rearranged into the form of ax² + bx = c, add both sides of the equation square the half the value of b.

  • [tex] ax^2 + bx + \Bigg( \frac{b}{2} \Bigg)^2 = c + \Bigg( \frac{b}{2} \Bigg)^2 \\ [/tex]

  • [tex] x^2 - 8x + \Bigg( \frac{-8}{2} \Bigg)^2 = -7 + \Bigg( \frac{-8}{2} \Bigg)^2 \\ [/tex]

  • [tex] x^2 - 8x + (- 4)^2 = -7 + (-4)^2 [/tex]

  • [tex] (x - 4)^2 = -7 + 16 [/tex]

  • [tex] (x - 4)^2 = 9 [/tex]

  • [tex] \sqrt{(x - 4)^2} = \pm \sqrt{9} [/tex]

  • [tex] x - 4 = \pm 3 [/tex]

  • [tex] x = \pm 3 + 4 [/tex]

• Solution 1:

  • [tex] x = 3 + 4 [/tex]

  • [tex] x = 7 [/tex]

• Solution 2:

  • [tex] x = - 3 + 4 [/tex]

  • [tex] x = 1 [/tex]

[tex] \large \therefore \underline{\boxed{\tt \purple{x = 7, \: x = 1}}} [/tex]

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