Question

फोर एक्स स्क्वायर प्लस 5 रूट 2 एक्स माइनस 3​

Answer 1

2453

Step-by-step explanation:

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Answer 2

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4x² +5root2x−3

=4xroot2 +6root2 x−root2x−3

=2root2x(root2x+3)−( root2 x+3)

=(root2 x+3)(2root2 x−1)

So, the zeros of the polynomial are

( root2x+3)=0 and (2root2 x−1)=0

x=-3root2/2 and x= 1/root2/2

x=-3root2/2 and x=root2/4

Now, the polynomial is 4x² +5root2x−3

Comparing it with ax²+bx+c=0, we get a=4, b=5root2and c=−3

Sum of the zeros =

=(-3root2/2)+(root4/2)

=-6root2+root2/4

=-5root2/4

=-b/a

Product of the zeros

=(-3root2/2)(root4/2)

=(-3/2)(2/4)

=(-3/4)

=c/a