Question

The sum of two numbers is 9 and their product is 20. Find the sum of their squares and cubes.​

Answer 1

Given :

• Sum of numbers = 9
• Product of numbers = 20

To Find :

• Sum of squares of these numbers
• Sum of cubes of these numbers

Solution :

• Let's the number x and y

Case 1 :

$\tt \implies x + y = 9 \\ \\ \tt \implies x = y - 9 \: \: \: \: \: \: \: ....(eq1)$

Case 2 :

$\implies\tt x \times y = 20$

Substitute value of x from (eq1)

$\implies\tt (9 - y) \times y= 20 \\ \\ \implies \tt - {y}^{2} + 9y - 20 = 0 \\ \\ \implies \tt {y}^{2} - 9y + 20 = 0\\ \\ \implies \tt {y}^{2} - 5y - 4y + 20 =0 \\ \\ \implies\tt y(y - 5) - 4(y - 5) = 0 \\ \\ \implies \tt (y - 5)(y - 4) = 0 \\ \\ \implies\tt y = 5 \: \: \: \: \: \: or \: \: \: \: \: \: y = 4$

Here y have two values we put y = 5 in (eq1) but we can also put y = 4

$\tt \implies x = 9 - y \\ \\ \implies \tt x = 9 - 5 \\ \\ \tt \implies x = 4$

now

1st number = 4

2nd number = 5

Sum of square of these numbers :

$\implies\tt {(4)}^{2} + {(5)}^{2} \\ \\ \implies\tt16 + 25 \\ \\ \implies\tt41$

Sum of cube of these numbers :

$\tt \implies {(4)}^{3} + {(5)}^{3} \\ \\ \tt \implies64 + 125 \\ \\ \tt \implies189$

Answer 2

Answer:

✅Verified answer✅

Sum of numbers = 9

Product of numbers = 20

To Find :

Sum of squares of these numbers

Sum of cubes of these numbers

Solution :

Let's the number x and y

Case 1 :

\begin{lgathered}\tt \implies x + y = 9 \\ \\ \tt \implies x = y - 9 \: \: \: \: \: \: \: ....(eq1)\end{lgathered}

⟹x+y=9

⟹x=y−9....(eq1)

Case 2 :

\implies\tt x \times y = 20⟹x×y=20

Substitute value of x from (eq1)

\begin{lgathered}\implies\tt (9 - y) \times y= 20 \\ \\ \implies \tt - {y}^{2} + 9y - 20 = 0 \\ \\ \implies \tt {y}^{2} - 9y + 20 = 0\\ \\ \implies \tt {y}^{2} - 5y - 4y + 20 =0 \\ \\ \implies\tt y(y - 5) - 4(y - 5) = 0 \\ \\ \implies \tt (y - 5)(y - 4) = 0 \\ \\ \implies\tt y = 5 \: \: \: \: \: \: or \: \: \: \: \: \: y = 4\end{lgathered}

⟹(9−y)×y=20

⟹−y

2

+9y−20=0

⟹y

2

−9y+20=0

⟹y

2

−5y−4y+20=0

⟹y(y−5)−4(y−5)=0

⟹(y−5)(y−4)=0

⟹y=5ory=4

Here y have two values we put y = 5 in (eq1) but we can also put y = 4

\begin{lgathered}\tt \implies x = 9 - y \\ \\ \implies \tt x = 9 - 5 \\ \\ \tt \implies x = 4\end{lgathered}

⟹x=9−y

⟹x=9−5

⟹x=4

now

1st number = 4

2nd number = 5

Sum of square of these numbers :

\begin{lgathered}\implies\tt {(4)}^{2} + {(5)}^{2} \\ \\ \implies\tt16 + 25 \\ \\ \implies\tt41\end{lgathered}

⟹(4)

2

+(5)

2

⟹16+25

⟹41

Sum of cube of these numbers :

\begin{lgathered}\tt \implies {(4)}^{3} + {(5)}^{3} \\ \\ \tt \implies64 + 125 \\ \\ \tt \implies189\end{lgathered}

⟹(4)

3

+(5)

3

⟹64+125

⟹189

Step-by-step explanation:

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