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Find the TWO integers whos product is -12 and whose sum is 1
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Sagot :

VirαtKσhliidn

Answer:

[tex] \rm Numbers = 4 \ and \ -3.[/tex]

Step-by-step explanation:

Given :-

The sum of two numbers is 1 .

The product of the nos . is 12 .

And we need to find out the numbers. So let us take ,

First number be x

Second number be 1-x .

According to first condition :-

[tex]\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}[/tex]

Hence the numbers are 4 and -3

Let integers be x and x-1

ATQ

  • x+x-1=1

[tex]\\ \sf\longmapsto 2x-1=1[/tex]

[tex]\\ \sf\longmapsto 2x=1+1[/tex]

[tex]\\ \sf\longmapsto 2x+2[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{2}{2}[/tex]

[tex]\\ \sf\longmapsto x=1[/tex]

Now

[tex]\\ \sf\longmapsto x-1=1-1=0[/tex]

  • The integers are 1 and 0

But

Integers can be x and 1-x as their sum is 1

[tex]\\ \sf\longmapsto x(1-x)=-12[/tex]

[tex]\\ \sf\longmapsto x- x^2=-12[/tex]

[tex]\\ \sf\longmapsto x^2-x-12=0[/tex]

[tex]\\ \sf\longmapsto x^2-4x+3x-12=0[/tex]

[tex]\\ \sf\longmapsto x(x-4)+3(x-4)=0[/tex]

[tex]\\ \sf\longmapsto (x-4)(x+3)=0[/tex]

[tex]\\ \sf\longmapsto x=4,-3[/tex]

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