if sin alfa and cos alfa are the roots of the equation ax2+bx+c=0, then provethat a2+2ac=b2 plz plz fast its my anual exam
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Answered by
3
let it be q rather than Alfa.becuz I have to write Alfa and it WL consume time.
so let the two roots be -b+√b²-4ac/2a and -b-√b²-4ac/2a.and we know that sin²q+cos²q=1.this prove this time taking.so I am doing it in short step.and if u solve this u wl get a²+2ac-b²=0.
so let the two roots be -b+√b²-4ac/2a and -b-√b²-4ac/2a.and we know that sin²q+cos²q=1.this prove this time taking.so I am doing it in short step.and if u solve this u wl get a²+2ac-b²=0.
Answered by
42
given quadratic equation ax²+bx+c=0
let one root = p = sin α
second root = q = cos α
i) sum of the roots = -b/a
p+q = - b/a
sin α+cos α = -b/a ---(1)
ii) pq = c/a
sin α cosα = c/a---(2)
do the square of (1)
sin² α +cos² α + 2 sin αcosα = b²/a²
1+2(c/a) = b²/a² from sin²α+cos²α =1 and (2)
multiply each term with a²
a² +2ac = b²
let one root = p = sin α
second root = q = cos α
i) sum of the roots = -b/a
p+q = - b/a
sin α+cos α = -b/a ---(1)
ii) pq = c/a
sin α cosα = c/a---(2)
do the square of (1)
sin² α +cos² α + 2 sin αcosα = b²/a²
1+2(c/a) = b²/a² from sin²α+cos²α =1 and (2)
multiply each term with a²
a² +2ac = b²
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