What 2 numbers make a product of 160 and sum of 12

first number is 20 Second number I -8

Step-by-step explanation:

Given \: sum \: of \: two \: numbers = 12Givensumoftwonumbers=12

Let \: One \: number = xLetOnenumber=x

Second \: number = ( 12 - x )Secondnumber=(12−x)

\blue { Product \: of \: two \: numbers = -160}Productoftwonumbers=−160

\implies x( 12 - x ) = -160⟹x(12−x)=−160

\implies 12x - x^{2} + 160 = 0⟹12x−x2+160=0

/* Multiplying bothsides by (-1) , we get */

\implies -12x + x^{2} - 160 = 0⟹−12x+x2−160=0

\implies x^{2} - 12x - 160 = 0⟹x2−12x−160=0

/* Splitting the middle term,we get */

\implies x^{2} + 8x - 20x - 160 = 0⟹x2+8x−20x−160=0

\implies x( x + 8 ) - 20( x + 8 ) = 0⟹x(x+8)−20(x+8)=0

\implies ( x + 8 )( x - 20 ) = 0⟹(x+8)(x−20)=0

\implies x + 8 = 0 \: Or \: x - 20 = 0⟹x+8=0Orx−20=0

\implies x = -8 \: Or \: x = 20⟹x=−8Orx=20

\begin{gathered} i ) If \: x = -8 \: then \\text{\O}ne \: number = -8 \end{gathered}i)Ifx=−8thentext\Onenumber=−8

\begin{gathered} Second \: number = 12 - x \\= 12 - (-8) \\= 12 + 8 \\= 20 \end{gathered}Secondnumber=12−x=12−(−8)=12+8=20

\begin{gathered} ii ) If \: x = 20 \: then \\text{\O}ne \: number = 20 \end{gathered}ii)Ifx=20thentext\Onenumber=20

\begin{gathered} Second \: number = 12 - 20 \\= -8 \end{gathered}Secondnumber=12−20=−8

Therefore.,

\begin{gathered} \red{ Required \: two \: numbers \: are } \\\green { -8 \: and \: 12 } \end{gathered}Requiredtwonumbersare−8and12

not possible, no products of 160 and to the sum of 12

Step-by-step explanation:

160, 1

80, 2

40, 4

32, 5

20, 8

16, 10

7,24

step-by-step explanation: