plzz answer thiss plz
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sweetylookss:
hello
the pic is soo clear only
jaa jii la apni zindgi
if u don't know the answer and snd like these stupid answers what will i doo
usko clear nhi all clear bolta ha
which class r u
answer delete ho gayaga
12
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1)only 2 but not 2√2...because by substituting √2...√2square+√2*√2-4=2+2-4=0
but 2√2 will be 2√2 square +2√2*√2-4= 8+4-4=8 which is not equal to zero.
2) substitute 2 in the equation. 5x^2+px-r=0
5*2^2+p*2-r=0
20+2p-r=0 let this equation be x
Substitute-1/5 in 5x^2+px-r=0
5(-1/5)^2+p(-1/5)-r=0
5(1/25)-p(1/5)-r=0
(1/5)-p(1/5)-r=0
take lcm
1-p-5r=0
p=1-5r..now substitute this in x
20+2(1-5r)-r=0
20+2-10r-r=0
r=22/11=2
now substitute r in p=1-5r
p=1-5(2)=-9
3)sub -a^2 in the equation...x^2+3ax+k=0
(-a)^2+3a(-a)+k=0
a^2-3a^2+k=o
k=3a^2-a^2=2a^2
4) idid not understand the question. I cannot see it properly
but 2√2 will be 2√2 square +2√2*√2-4= 8+4-4=8 which is not equal to zero.
2) substitute 2 in the equation. 5x^2+px-r=0
5*2^2+p*2-r=0
20+2p-r=0 let this equation be x
Substitute-1/5 in 5x^2+px-r=0
5(-1/5)^2+p(-1/5)-r=0
5(1/25)-p(1/5)-r=0
(1/5)-p(1/5)-r=0
take lcm
1-p-5r=0
p=1-5r..now substitute this in x
20+2(1-5r)-r=0
20+2-10r-r=0
r=22/11=2
now substitute r in p=1-5r
p=1-5(2)=-9
3)sub -a^2 in the equation...x^2+3ax+k=0
(-a)^2+3a(-a)+k=0
a^2-3a^2+k=o
k=3a^2-a^2=2a^2
4) idid not understand the question. I cannot see it properly
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