(PLEASE HELP!!! DUE 4:30!!!) Image of the problem is attached.
1 answer:
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Find the roots
solve
we use hmm, completing the suare
2(x²-1.5x)=4
divide both sides by 2
x²-1.5x=2
take 1/2 of linear coeiftn and square it
-1.5/2=-0.75, (-0.75)²=0.5625
add that to both sides
x²-1.5x+0.5625=2+0.5625
factor perfect squaer trinomial
(x-0.75)²=2.5625
square root both sides, remember to take positive and negative square roots
x-0.75=+/-√2.5625
add 0.75 to both sides
x=0.75+/-√2.5625
the roots are x=0.75+√2.5625 and x=0.75-√2.5625
1/a and 1/b
1/(0.75+√2.5625) and 1/(0.75-√2.5625)
if the roots of a quadratic equation are r1 and r2 then it factors to
(x-r1)(x-r2)
so then we can factor our equation to be
if we were to try and expand it, we would get
x²+0.75x-0.5
that's the simpliest equation with roots 1/a and 1/b where a and b are he roots of 2x²-3x=4
x²+0.75x-0.5 is answer
solve
we use hmm, completing the suare
2(x²-1.5x)=4
divide both sides by 2
x²-1.5x=2
take 1/2 of linear coeiftn and square it
-1.5/2=-0.75, (-0.75)²=0.5625
add that to both sides
x²-1.5x+0.5625=2+0.5625
factor perfect squaer trinomial
(x-0.75)²=2.5625
square root both sides, remember to take positive and negative square roots
x-0.75=+/-√2.5625
add 0.75 to both sides
x=0.75+/-√2.5625
the roots are x=0.75+√2.5625 and x=0.75-√2.5625
1/a and 1/b
1/(0.75+√2.5625) and 1/(0.75-√2.5625)
if the roots of a quadratic equation are r1 and r2 then it factors to
(x-r1)(x-r2)
so then we can factor our equation to be
if we were to try and expand it, we would get
x²+0.75x-0.5
that's the simpliest equation with roots 1/a and 1/b where a and b are he roots of 2x²-3x=4
x²+0.75x-0.5 is answer