Question

QUADRATIC EQUATO W ON
To work on
The difference of two positive whole numbers is 3 and the sum of their squares is 117; by
calculating, let us write the two numbers.
at all​

Answer 1

Answer:

Required two numbers are

Required two numbers are 9,6

Step-by-step explanation:

Let x,(x-3) are two positive whole numbers.

According to the problem given,

x²+(x-3)²=117

\implies x^{2}+x^{2}-2\times x\times 3+3^{2}=117⟹x

2

+x

2

−2×x×3+3

2

=117

\implies 2x^{2}-6x+9-117=0⟹2x

2

−6x+9−117=0

\implies 2x^{2}-6x-108=0⟹2x

2

−6x−108=0

/* Divide each term by 2, we get

\implies x^{2}-3x-54=0⟹x

2

−3x−54=0

/* Splitting the middle term, we get

\implies x^{2}-9x+6x-54=0⟹x

2

−9x+6x−54=0

\implies x(x-9)+6(x-9)=0⟹x(x−9)+6(x−9)=0

\implies (x-9)(x+6)=0⟹(x−9)(x+6)=0

\implies x-9=0 \: Or \:x+6=0⟹x−9=0Orx+6=0

\implies x=9 \: Or \:x=-6⟹x=9Orx=−6

But , x is a whole number.

x = 9

Required two numbers are x,x-3

9,6