Question

the product of two integers is -144. and sim of the integer is -7 . find the smallest number. Please explain the answer​

Answer 1

explanation :

let two integers be p, q

given pq= -144

p+q = -7

If p+q and pq are known then quadratic equation corresponding to roots as p and qis given by,

x^2−(p+q)x+pq=0

x^2-(-7)x+(-144)=0

x^2+7x-144=0

x^2+16x-9x-144=0

x(x+16)-9(x+16)=0

(x-9)(x+16)=0

x-9=0 x+16=0

x=9,x=-16

therfore numbers are 9,-16

Answer 2

Concept:-

Here, We have given the product of two integers and the sum of two integers. We have to find the smallest integer in both integers. Let's find

Given:-

• Product of two integers is -144.
• Sum of two integers is -7.

To Find:-

• Smallest number in both the integers ?

Solution:-

Here, Let the one integer be x.

Then, Another integer will be y.

Now, According to the question we have,

$\sf{xy = -144}$      -------------- (i)

$\sf{x + y = -7}$     --------------- (ii)

By solving equation (ii) we get,

$\sf{x = -7 - y}$   ----------- (iii)

Now, Putting the equation (iii) in equation (i) we get,

$\sf{\Longrightarrow (-7 - y)y = -144}$

$\sf{\Longrightarrow -(7y + y^{2}) = -144}$                           [ Taking (-) common ]

$\sf{\Longrightarrow y^{2} + 7y = 144}$

$\sf{\Longrightarrow y^{2} + 7y - 144 = 0}$

By middle term splitting,

$\sf{\Longrightarrow y^{2} + 16y - 9y - 144 = 0}$

$\sf{\Longrightarrow y(y + 16) - 9(y + 16) = 0}$

$\sf{\Longrightarrow (y + 16) (y - 9) = 0}$

Equating both the equations separately we get,

$\Longrightarrow \begin{array}{c|c} \bf y - 9 = 0 & \bf y + 16 = 0\end{array}$

$\Longrightarrow \begin{array}{c|c} \bf y = 9 & \bf y = -16\end{array}$

Hence, The smallest number is -16.