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QUADRATIC EQUATO W ONTo work onThe difference of two positive whole numbers is 3 and the sum of their squares is 117; bycalculating, let us write the two numbers.at all |
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Answer» 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 |
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