Question

The product of two no.'s is 192 and the su, of these no.'s is 28 . Find the no.'s..​

Answer 1

✴Ola!!✴

⤵⤵Answer⤵⤵

Let the required numbers be x and (28-x).

Then,

x(28-x)=192 <=> 28x-x^2 = 192

<=> x^2 - 28x +192 = 0

<=> x^2-16x-12x+192 = 0

<=> x(x-16)-12(x-16) = 0

<=> (x-16)(x-12) = 0

<=> x-16 = 0 or x-12 = 0

<=> x = 16 or x = 12

Hence, the required numbers are 16 and 12 .✔✔✔

Hope it helps u✌✌

tysm❤❤

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@avinam27✌✌

Answer 2

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⭕According to QUESTION I THINK Firstly we have to assume two no.

So, Let the →First No. be 'x' and Second No. be → '28-x' [as sum of no. is given 28]

Now , let see what QUESTION is saying...

⚫ Product (means multiple) of both the no. is 192 ...

i.e., x(28-x) = 192_________[equ.(i)]

⚫Sum (means add) of both no. is 28

i.e., x+(28-x) = 28_________[equ.(ii)]

☑️Now We can find both no. by evaluating any one of equation..

So, Now by evaluating equ.(i)

=> x(28-x) = 192

=> 28x-x²= 192

=> x²-28x+192 = 0 ............{by taking the equation of L.H.S into side of R.H.S in order to make equation +ve}

⚫By mid Term factorization..

=> x²-16x-12x+192 = 0

=> x(x-16) - 12(x-16) = 0 ...........{taking X as common}

=> (x-16) (x-12)

From here we get values of X i.e., 16 and 12

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⚫As we assumed firstly that First no. be 'x' So, the By getting value of x we can put any one value (both are applicable) = 16/12..

and the Second no. as we assumed (28-x) .

= (28-16) = 12 or, (28-12) = 16....

☑️HENCE☑️WE GOT THE ANSWER ☝️☝️