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anonymous

  • one year ago

How many real solutions does each quadratic equation shown below have?

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  1. anonymous
    • one year ago
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    \[x^{2}+\left( \frac{ 4 }{ 5 }\right)x =-1/4\]

  2. Jhannybean
    • one year ago
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    First we need to put our quadratic in the proper quadratic form. \[ax^2+bx+c=0\] To do that we first retrice\(+\dfrac{1}{4}\) to both sides of the equation.

  3. anonymous
    • one year ago
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    \[x^{2}-7x +10=0\]

  4. anonymous
    • one year ago
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    That's the second equation

  5. Jhannybean
    • one year ago
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    Alright, for the first one did you set it up in the proper quadratic form?

  6. Owlcoffee
    • one year ago
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    Any quadratic equation, with the form: \(ax^2 \pm bx \pm c=0\) present two, one, or no solution depending on the value of a component called the "discriminant", the discriminant is a component of the solution, or more specifically, the general formula that allows us to know with little operations if a quadratic equation has two, on or no solution. The discriminand, often notated with the greek letter "delta" is composed by the values inside the square root of the general formula: \[\Delta = b^2 - 4ac\] So, if \(\Delta > 0\) the quadratic equation presents two solutions \(x_1\) and \(x_2\). if \(\Delta = 0\) : The quadratic equation presents only one solution x. if \(\Delta<0\): The quadratic equation presents no solution inside the real numbers.

  7. anonymous
    • one year ago
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    \[(4/5)^{2}-4(1)(1/4)=16/25-1<0\]so there's no real solutions for this one

  8. Jhannybean
    • one year ago
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    You should have \[x^2+\frac{4}{5}x+\frac{1}{4}=0\] Then you identify what \[a=\]\[b=\]\[c=\] The determinant of the quadratic function, if you did not know already, is represented by the form, \(b^2-4ac\). Plugging in the values of a b and c, we can tell if it will have 2 real solutions if the determinant is positive, \[b^2-4ac > 0\] 0 solution if its negative, \[b^2-4ac <0\] and 1 solution if it is = 0 \[b^2-4ac = 0\]

  9. Jhannybean
    • one year ago
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    That's correct.

  10. Jhannybean
    • one year ago
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    Now how about your second function, \(x^{2}-7x +10=0\)?

  11. anonymous
    • one year ago
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    I found two solutions for the second one

  12. Jhannybean
    • one year ago
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    You've got \(a=1~,~ b=-7~,~ c=10\) \[b^2-4ac = (-7)^2 -4(1)(10) = 9 > 0 \]

  13. Jhannybean
    • one year ago
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    Yep, I did too.

  14. anonymous
    • one year ago
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    Yay!!! the last equation is \[x ^{2}-(\frac{ 2 }{ 3 })x +\frac{ 1 }{ 9 }=0\]

  15. Jhannybean
    • one year ago
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    So what does a b and c = ?

  16. Jhannybean
    • one year ago
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    Remember, your quadratic is in the form: \[\color{red}{a}x^2+\color{blue}{b}x+\color{green}{c} = 0\]\[~~~~~~~~~~\downarrow\]\[\color{red}{1}x^2+\left(\color{blue}{-\frac{2}{3}}\right)x+\color{green}{\frac{1}{9}}=0\]

  17. anonymous
    • one year ago
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    I got 1 solution

  18. Jhannybean
    • one year ago
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    Me too.

  19. anonymous
    • one year ago
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    Yay!!!

  20. Jhannybean
    • one year ago
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    Good job.

  21. anonymous
    • one year ago
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    Thanks :)

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