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anonymous

  • one year ago

x^2+5x-24 = 0 + 24 +24 x^2+5x=24 Then I know you square root both sides but that's when I get stuck, I'm thinking it would then be something like x + ? = 2sqrt6 The answer is -5 so I just need help being taught how to do these last steps, unless I'm completely wrong in the beginning steps.

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  1. alekos
    • one year ago
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    Are you attempting to solve for x using the complete the square method?

  2. anonymous
    • one year ago
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    I'm trying to get the sum of 2 solutions of the equation \[X^{2}+5x-24=0\]

  3. alekos
    • one year ago
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    you mean the product of two solutions, in other words you want to factorise the quadratic

  4. alekos
    • one year ago
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    (x+a)(x+b) = x^2 +5x -24

  5. mathstudent55
    • one year ago
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    Solve the equation using the complete the square method. Then add the two solutions together. The final answer is indeed -5. Now just explain how to do it.

  6. alekos
    • one year ago
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    Hmm. I don't get -5 as one of the solutions?

  7. mathstudent55
    • one year ago
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    Step 1. Let's solve the quadratic equation using the complete the square method. \(x^2+5x-24 ~~~= ~~~0\) \(~~~~~~~~~~~~+24~~~~+24\) -------------------- \(x^2 + 5x ~~~~~~~~~~=~~24\) The complete the square step is to add the square of half of the x-term coefficient. The x-term coefficient is 5. Half of 5 is \(\dfrac{5}{2} \). The square of \(\dfrac{5}{2} \) is \(\dfrac{25}{4} \). We add \(\dfrac{25}{4} \) to both sides to complete the square.

  8. mathstudent55
    • one year ago
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    \(x^2 + 5x + \dfrac{25}{4} = 24 + \dfrac{25}{4} \) \(x^2 + 5x + \dfrac{25}{4} = \dfrac{96}{4} + \dfrac{25}{4} \) \(x^2 + 5x + \dfrac{25}{4} = \dfrac{121}{4}\) \(\left( x + \dfrac{5}{2} \right)^2 = \dfrac{121}{4} \) We have finished the step of completing the square. Now we take the square root of both sides.

  9. mathstudent55
    • one year ago
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    Step 2. Take square root of both sides. Rule: If \(x^2 = k\), where \(k\) is a number, then \(x = \pm k\). Because of the rule above, we must be careful to include both the positive and negative roots of the number on the right side of the equal sign.

  10. mathstudent55
    • one year ago
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    \(\left( x + \dfrac{5}{2} \right)^2 = \dfrac{121}{4}\) \( x + \dfrac{5}{2} = \pm\sqrt{\dfrac{121}{4}}\) Notice the \(\pm\) sign on the right side. Step 3. Separate the equations and solve for x. \( x + \dfrac{5}{2} = \pm \dfrac{11}{2}\) Now we separate the above into two equations, one for the positive root, and one for the negative root. \(x + \dfrac{5}{2} = \dfrac{11}{2}\) or \(x + \dfrac{5}{2} = -\dfrac{11}{2}\) \(x = \dfrac{6}{2}\) or \(x = -\dfrac{16}{2}\) \(x = 3\) or \(x = -8\) The solutions to the quadratic equation are 3 and -8. When we add the solutions, we get: \(3 + (-8) = -5\)

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