Respuesta :
Step-by-step explanation:
It is given that, the sum of two positive integers is 31 and the sum of the squares of these numbers is 625 and we are to find the smaller of the numbers.
So, let the two positive integers be x and y.
Therefore,
[tex] \\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 \: \: ........ \: (i) }}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + y {}^{2} = 625 \: ...... \: (ii)}}} \\ \\[/tex]
Now, From the first equation we have,
[tex]\\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 }}} \\ \\ [/tex]
[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - x \: ...... \: (iii)}}} \\ \\ [/tex]
Now, substituting the value of y in equation (ii) we get :
[tex] \\ {\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 - x) {}^{2} = 625}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 {}^{2} - 2.x.31 + x{}^{2} ) = 625}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (961 - 62x + x{}^{2} ) = 625}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + 961 - 62x + x{}^{2} - 625 = 0}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} + 336 - 62x = 0}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} - 62x + 336 = 0}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad 2( x {}^{2} - 31 + 168) = 0}}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad \frac{2}{2} ( x {}^{2} - 31 + 168) = \frac{0}{2} }}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} - 31 + 168 = 0}}} \\ \\[/tex]
Now using the quadratic formula :
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}} \\ \\[/tex]
Where,
- a = 1
- b = -31
- c = 168
[tex] \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - (- 31) \pm \sqrt{ {31}^{2} - 4(1)(168)} }{2(1)} }}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 961 -672} }{2(1)} }}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 289} }{2(1)} }}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm 17 }{2(1)} }}} \\ \\[/tex]
Now, we have two equations,
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} \: ... .....\: (iv)}}} \\ \\[/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2} ... .....\: (v)}}} \\ \\[/tex]
So, Equation (iv) :
[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} }}} \\ \\ [/tex]
[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 48 }{2} }}} \\ \\ [/tex]
[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = 24 }}} \\ \\ [/tex]
Now, Equation (v) :
[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2}}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 14 }{2}}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad x = \: 7}}}\\ \\ [/tex]
- So, the value of x is 7 or 24
Now, we are to find the value of y.
Substituting the value of x (24) in equation (iii) :
[tex] \\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad y = 31 - 24 \:}}} \\ \\ [/tex]
[tex]{\longrightarrow \pmb{\sf {\qquad y = 7 \:}}} \\ \\ [/tex]
Again, Substituting the value of x (7) in equation (iii) :
[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]
[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - 7 \:}}} \\ \\ [/tex]
[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 24 \:}}} \\ \\ [/tex]
Therefore,
- The value of y is also 7 or 24.
[tex] \\ [/tex]
So, The smaller of the numbers is 7 .
Question : -
The sum of two positive integers is 31. If the sum of the squares of these numbers is 625, find the smaller of the numbers.
Given : -
- Sum of two positive numbers = 31
- Sum of squares of these numbers = 625
To Find : -
- We have to find the smaller of the numbers .
Concept : -
This question belongs to quadratic equations so we have to find the answer by making equation and solving it .
To Assume : -
- Let the first no. be x
- Let the second no. be y
So let's get started with Solution :
According to question , sum of two positive integers is 31 . So ,
- x + y = 31 --------- ( Equation 1 )
According to question , sum of square
of these numbers is 625 . So ,
- x² + y² = 625 --------- ( Equation 2 )
From equation 1 ( x + y = 31 ) , Value
of x :
- x = 31 - y
Now , putting value of x in eq. 2 :
- x² + ( 31 - x )² = 625
- x² + ( 31 )² - ( 2 × 31 × x ) + x² = 625
- 2x² + 961 - 62x = 625
- 2x² - 62x = 625 - 961
- 2x² - 62x = -336
- 2x² - 62x + 336 = 0
- 2(x² - 31x + 168 ) = 0
- x² - 31x + 168 = 0
Solving it by using middle term
splitting :
- x² -24x -7x + 168 = 0
- x ( x - 24 ) -7 ( x - 24 ) = 0
- ( x - 7 ) ( x - 24 )
So ,
First number ,
- x - 7 = 0
- x = 7 { Smaller Number }
Second Number ,
- x - 24 = 0
- x = 24
Verification :
According to question ,
Sum of numbers is equal to 31 :
- x + y = 31
- 7 + 24 = 31
- L.H.S = R.H.S
Sum of squares of these numbers is equal to 625 :
- x² + y² = 625
- 7² + 24² = 625
- 49 + 576 = 625
- 625 = 625
- L.H.S = R.H.S
