Question

If (2p + 3q = 10) and (8p3 + 27q3 = 100), find the value of pq.

Given that
2p+3q=102p+3q=10 ...[Equation 1]
8p^3+27q^3=1008p3+27q3=100 ...[Equation 2]
Factorization of Equation 2
\rightarrow (2p+3q)(4p^2-6pq+9p^2)=100→(2p+3q)(4p2−6pq+9p2)=100
\rightarrow 10(4p^2-6pq+9q^2)=100→10(4p2−6pq+9q2)=100
\rightarrow 4p^2-6pq+9q^2=10→4p2−6pq+9q2=10 ...[Equation 3]
Since this question requires identities, for convenience, let the sum and product of 2p,3q2p,3q be m,nm,n .
Then Equation 1, 3 are
\begin{gathered}\rightarrow \begin{cases} & m=10 \\ & m^2-3n=10 \end{cases}\end{gathered}→{​m=10m2−3n=10​​
\rightarrow m=10,n=30→m=10,n=30
Hence 6pq=306pq=30 , and hence pq=5pq=5 .



Hope it helps=)​

Answer 1

Answer:

2p+3q=102p+3q=10 ...[Equation 1]

8p^3+27q^3=1008p3+27q3=100 ...[Equation 2]

Factorization of Equation 2

\rightarrow (2p+3q)(4p^2-6pq+9p^2)=100→(2p+3q)(4p2−6pq+9p2)=100

\rightarrow 10(4p^2-6pq+9q^2)=100→10(4p2−6pq+9q2)=100

\rightarrow 4p^2-6pq+9q^2=10→4p2−6pq+9q2=10 ...[Equation 3]

Since this question requires identities, for convenience, let the sum and product of 2p,3q2p,3q be m,nm,n .

Then Equation 1, 3 are

\begin{gathered}\rightarrow \begin{cases} & m=10 \\ & m^2-3n=10 \end{cases}\end{gathered}→{m=10m2−3n=10

\rightarrow m=10,n=30→m=10,n=30

Hence 6pq=306pq=30 , and hence pq=5pq=5 .

Step-by-step explanation:

hope it helps you