find the value:-
where a+b=8
and. ab=15
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Answered by
16
By hit and trial we get 3 & 5.
By solving,
a+b=8=>a2+2ab+b2=64.(i)
ab=15=>4ab=60.(ii)
Subtracting (ii) from (i),
a2-2ab+b2=4=>a-b=±2(iii)
From (iii), we get:
a-b=2(iv)&a-b=-2(v)
From (i),a+b=8.
From(iv),a-b=2
Adding them we get,
2a=10=>a=5.(vi)
From (i)&(vi), we get
5+b=8=>b=3
So, first set(a,b)=(5,3)
Now, from(v) a-b=-2
Adding(i)&(v),
2a=6=>a=3(vii)
From(i)&(vii), we get
3+b=8=>b=5.
So,(a,b)=(3,5)
Therefore (a,b)=(5,3)&(3,5).
By solving,
a+b=8=>a2+2ab+b2=64.(i)
ab=15=>4ab=60.(ii)
Subtracting (ii) from (i),
a2-2ab+b2=4=>a-b=±2(iii)
From (iii), we get:
a-b=2(iv)&a-b=-2(v)
From (i),a+b=8.
From(iv),a-b=2
Adding them we get,
2a=10=>a=5.(vi)
From (i)&(vi), we get
5+b=8=>b=3
So, first set(a,b)=(5,3)
Now, from(v) a-b=-2
Adding(i)&(v),
2a=6=>a=3(vii)
From(i)&(vii), we get
3+b=8=>b=5.
So,(a,b)=(3,5)
Therefore (a,b)=(5,3)&(3,5).
uneq95:
bro, you have to find the value of expression
I just did not see the picture. Anyway, sice I have given the values of a,b, he can now find it out .
I meant since not sice
but that was totally unnecessary
Actually, I did not see the pic.Your answer is more simple and accurate.
ok
Answered by
26
Actually the approach to solve the problem is given in the hint.
a⁴+a²b²+b⁴ = (a²+b²)² - a²b²
Now, (a² + b²) = (a+b)²-2ab
{(a² + b²)}² = {(a+b)²-2ab}²
a⁴+a²b²+b⁴ ={(a+b)²-2ab}² - a²b²
Now , since a+b = 8
ab=15
Just substitute these values and you will get the answer.
a⁴+a²b²+b⁴ ={(8)²-2×15}² - (15)²
= {64-30}²-225 = 34²-225= 1156-225= 931
a⁴+a²b²+b⁴ = (a²+b²)² - a²b²
Now, (a² + b²) = (a+b)²-2ab
{(a² + b²)}² = {(a+b)²-2ab}²
a⁴+a²b²+b⁴ ={(a+b)²-2ab}² - a²b²
Now , since a+b = 8
ab=15
Just substitute these values and you will get the answer.
a⁴+a²b²+b⁴ ={(8)²-2×15}² - (15)²
= {64-30}²-225 = 34²-225= 1156-225= 931
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