Question

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

Answer 1

\[x^{2}+\left( \frac{ 4 }{ 5 }\right)x =-1/4\]

Answer 2

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.

Answer 3

\[x^{2}-7x +10=0\]

Answer 4

That's the second equation

Answer 5

Alright, for the first one did you set it up in the proper quadratic form?

Answer 6

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.

Answer 7

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

Answer 8

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 \[b^2-4ac = 0\]

Answer 9

That's correct.

Answer 10

Now how about your second function, \(x^{2}-7x +10=0\)?

Answer 11

I found two solutions for the second one

Answer 12

You've got \(a=1~,~ b=-7~,~ c=10\)

\[b^2-4ac = (-7)^2 -4(1)(10) = 9 > 0 \]

Answer 13

Yep, I did too.

Answer 14

Yay!!!
the last equation is
\[x ^{2}-(\frac{ 2 }{ 3 })x +\frac{ 1 }{ 9 }=0\]

Answer 15

So what does a b and c = ?

Answer 16

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

Answer 17

I got 1 solution

Answer 18

Me too.

Answer 19

Yay!!!

Answer 20

Good job.

Answer 21

Thanks :)