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anonymous
 one year ago
How many real solutions does each quadratic equation shown below have?
anonymous
 one year ago
How many real solutions does each quadratic equation shown below have?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^{2}+\left( \frac{ 4 }{ 5 }\right)x =1/4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's the second equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, for the first one did you set it up in the proper quadratic form?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0Any 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(4/5)^{2}4(1)(1/4)=16/251<0\]so there's no real solutions for this one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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, \(b^24ac\). Plugging in the values of a b and c, we can tell if it will have 2 real solutions if the determinant is positive, \[b^24ac > 0\] 0 solution if its negative, \[b^24ac <0\] and 1 solution if it is = 0 \[b^24ac = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now how about your second function, \(x^{2}7x +10=0\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found two solutions for the second one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You've got \(a=1~,~ b=7~,~ c=10\) \[b^24ac = (7)^2 4(1)(10) = 9 > 0 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yay!!! the last equation is \[x ^{2}(\frac{ 2 }{ 3 })x +\frac{ 1 }{ 9 }=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what does a b and c = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember, your quadratic is in the form: \[\color{red}{a}x^2+\color{blue}{b}x+\color{green}{c} = 0\]\[~~~~~~~~~~\downarrow\]\[\color{red}{1}x^2+\left(\color{blue}{\frac{2}{3}}\right)x+\color{green}{\frac{1}{9}}=0\]
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