Question

find the zeroes of the quadratic polynomial:
a(x²+1)-x(a²+1)
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Answer 1

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Answer 2

Answer:

a(x²+1)-x(a²+1)

=ax²+a - xa²-x

then we take

ax²+a - xa²-x =0

ax²- xa² -x +a=0

taking 'a' as common

a ( x²-xa+1) -x =0

then,  (x²-xa+1) -x =0×a=0

x²-ax+(1-x)=0

now we have a polynomial of degree 2

we can now find x using formula for solution of a quadratic equation

(-b±√b²-4ac²)÷2a

here, a=1      b= -a      c= 1-x

so,    [-(-a)±√(-a)²-4×1×(1-x) ] ÷ 2×1

so the two roots are

$\frac{a+\sqrt{a^{2} -4+4x} }{2a}$

$\frac{a-\sqrt{a^{2} -4+4x} }{2a}$

hope it helps..........