## Diana.xL one year ago help?

1. Diana.xL

Find the value of the discriminant for each quadratic equation below. Show all steps needed to write the answer in simplest form, including substituting the values of a, b, and c in the discriminant formula. Then use the value to determine how many real number solutions each equation has. 5x^2 + x = 4

2. misty1212

HI!!

3. Diana.xL

Hey :)

4. misty1212

start by setting it equal to zero go from $5x^2 + x = 4$ to $5x^2+x-4=0$

5. misty1212

$\large \color{red}ax^2+\color{blue}bx+\color{green}c=0$ $\huge \color{red}5x^2+\color{blue}1x+\color{green}{-4}=0$

6. misty1212

then as @risn said compute $\huge \color{blue}b^2-4\times \color{red}a\times \color{green}c$

7. Diana.xL

D= 1 - 4 (1)(-4)?

8. misty1212

nope

9. misty1212

$$\color{red}a=\color{red}{5}$$

10. Diana.xL

oh sorry. D = 1-4(5)(-4)

11. misty1212

yeah that is right what do you get?

12. Diana.xL

81?

13. Diana.xL

14. misty1212

yeah me too

15. Diana.xL

at the end of the question it asks us "Then use the value to determine how many real number solutions each equation has".

16. misty1212

that is the answer to Show all steps needed to write the answer in simplest form, including substituting the values of a, b, and c in the discriminant formula.

17. misty1212

ok two steps here 81 is positive, so there are two real solutions also 81 is a "perfect square" it is the square of 9 that means both solutions are rational numbers

18. Diana.xL

oh ok! thank u so much! :)

19. misty1212

in fact that means $5x^2+x-4=0$ can be solved by factoring you get $(5x-4)(x+1)=0$ so $x=\frac{4}{5}$ or $x=-1$

20. misty1212

$\color\magenta\heartsuit$