Tristaidn
Answered

IDNStudy.com, kung saan nagtatagpo ang mga eksperto para sagutin ang iyong mga tanong. Sumali sa aming komunidad ng mga bihasa upang makahanap ng mga sagot na kailangan mo sa anumang paksa o problema.

s²+4s-21=0

4x²-32x=28

r²-10r=-17

m²+7m-[tex] \frac{51}{9} [/tex]=0

please show your solutions.i need it right now.im so stress because of that.HUHUHU.


Sagot :

[tex]s^2+4s-21=0[/tex]
[tex]s^2+4s=21[/tex]
[tex]s^2+4s+(2)^2=21+(2)^2[/tex]
[tex] \sqrt{(s+2)^2}= \sqrt{25} [/tex]
[tex]s+2=5[/tex]
[tex]s=5-2[/tex]
[tex]s=3[/tex]

[tex]4x^2-32x=28[/tex]
[tex] \frac{4x^2-32x}{4}= \frac{28}{4} [/tex]
[tex]x^2-8x+(4)^2=7+(4)^2[/tex][tex] \sqrt{(x-4)^2} = \sqrt{23} [/tex]
[tex]x-4= \sqrt{23} [/tex]
[tex]x= \sqrt{23}+4 [/tex]
[tex]x=- \sqrt{23}+4 [/tex]

[tex]r^2-10r=-17[/tex]
[tex]r^2-10r+(5)^2=-17+(5)^2[/tex]
[tex] \sqrt{(r-5)^2}= \sqrt{8} [/tex]
[tex]r-5= \sqrt{8} [/tex]
[tex]r= \sqrt{8}+5 [/tex]
[tex]r=- \sqrt{8}+5 [/tex]

[tex]m^2+7m- \frac{51}{9}=0 [/tex]
[tex]m^2+7m= \frac{51}{9} [/tex]
[tex]m^2+7m+( \frac{7}{2})^2= \frac{51}{9}+( \frac{7}{2})^2 [/tex]
[tex] \sqrt{(m+ \frac{7}{2})^2 }= \sqrt{ \frac{215}{12} } [/tex]
[tex]m+ \frac{7}{2} = \sqrt{ \frac{215}{12} } [/tex]
[tex]m= \sqrt{ \frac{215}{12}[/tex][tex]- \frac{7}{2} [/tex]
[tex]m= \sqrt{ \frac{215}{12}} -\frac{42}{12} [/tex]
[tex]m=- \sqrt{ \frac{215}{12} }- \frac{42}{12} [/tex]

Phewww! That was hard but fun! :D Hope that helps ^_^
s² + 4s - 21 = 0
   (s+7) (s-3) = 0
s+7=0  ;  s-3=0
s = -7  ;  s = 3


4x² - 32x = 28
x² - 8x = 7
x² - 8x + 16 = 7 + 16
(x-4)² = 23
√(x-4)² = √23
x - 4 = ±√23
x = ±√23 +4
x = √23 +4    ;   x = -√23 +4


r² - 10r = -17
r² - 10r + 25 = -17 + 25
(r-5)² = 8
√(r-5)² = √8
r-5 = ±√8
r = ±2√2 +5
r = 2√2 +5   ;   r = -2√2 +5


m² +7m - 51/9 = 0
m² +7m = 17/3
m² +7m + (7/2}² = 17/3 + (7/2}²
√(m+7/2)² = √(215/12)
m + 7/2 = ± √(645)/ 6
m = [tex] \frac{ \sqrt{645} }{6} + \frac{7}{2} [/tex]    ;    m =  -\frac{ \sqrt{645} }{6} +  \frac{7}{2}