Question

Activity 1.2: Who I Am?
Directions: Categorize the mathematical sentences below using the given table. U
a separate sheet of
paper
3x2 + x + 10 > 0
x² - 4x+8 = 0
(2n2 + 5)(3n - 1) > 0
m + 5 S-12
5 + 6k = 7k2
(2r - 3)(r + 1) 22
7d2 s 14
2s2 - 3s + 5 < 0​

Answer 1

Answer:

very very very make me brainlist

Answer 2

Answer:

3x² + x + 10 > 0 has no real roots

x² - 4x+8 = 0 has no real roots

5 + 6k = 7k2 , k = $\frac{3 +- 2\sqrt{11 } }{7}$

2s2 - 3s + 5 < 0​ has no real roots

Step-by-step explanation:

3x² + x + 10 > 0

discriminant (d) = $b^2 -4ac$

here a = 3, b = 1 ,c =10

d= $1^2 - 4* 3*10$

d = -119

d<0

discriminant  is negative hence no real roots

x² - 4x+8 = 0

discriminant (d) = $b^2 -4ac$

here b = -4 , a = 1 ,c = 8

d = $4^2 - 4* 1 *8$

d= -16  

d < 0

discriminant  is negative hence no real roots

(2n2 + 5)(3n - 1) > 0

2$n^2$ + 5 > 0

$n^2$ > -5/2

square of any number is always greater than -5/2

hence it is true for all n

3n-1>0

n>1/3

final sol => n >1/3

5 + 6k = 7k2

7$k^2$ - 6k -5 = 0

Use the quadratic formula

k =  $\frac{-b +- \sqrt{b^2 -4ac} }{2a}$

here a = 7 , b = -6 , c = -5

k = $\frac{-(-6) +- \sqrt{(-6)^2 -4.7.(-5)} }{2.7}$

k =$\frac{6 +- \sqrt{36 +140 } }{14}$

k = $\frac{6 +- \sqrt{176 } }{14}$

k = $\frac{6 +- 4\sqrt{11 } }{14}$

k = $\frac{3 +- 2\sqrt{11 } }{7}$

2s2 - 3s + 5 < 0​

discriminant (d) = $b^2 -4ac$

here b = -3 , a = 2,c = 5

d = $3^2 - 4* 2 *5$

d= -31

d < 0

discriminant  is negative hence no real roots

#SPJ3