Tristaidn
Answered

IDNStudy.com, kung saan nagtatagpo ang mga eksperto para sagutin ang iyong mga tanong. Alamin ang mga maaasahang sagot sa iyong mga tanong mula sa aming malawak na kaalaman sa mga eksperto.

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}