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

1. anonymous

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

2. anonymous

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

$x^{2}-7x +10=0$

4. anonymous

That's the second equation

5. anonymous

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

6. Owlcoffee

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

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

8. anonymous

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. anonymous

That's correct.

10. anonymous

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

11. anonymous

I found two solutions for the second one

12. anonymous

You've got $$a=1~,~ b=-7~,~ c=10$$ $b^2-4ac = (-7)^2 -4(1)(10) = 9 > 0$

13. anonymous

Yep, I did too.

14. anonymous

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

15. anonymous

So what does a b and c = ?

16. anonymous

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

I got 1 solution

18. anonymous

Me too.

19. anonymous

Yay!!!

20. anonymous

Good job.

21. anonymous

Thanks :)