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Knowing that: a+b=5 and ab=2

Calculate: a(to the power of 4) + b(to the power of 4)

Provide complete procedure

Sagot :

Answer:

a⁴ + b⁴ = 433

Step-by-step explanation:

[tex]$\begin{aligned}(a+b)^4 & =a^4+4 a^3 b+6 a^2 b^2+4 a b^3+b^4 \\ & =a^4+b^4+6(a b)^2+a^2 \times 4 a b+b^2 \times 4 a b \\ & =a^4+b^4+6(a b)^2+4 a b \times\left(a^2+b^2\right) \\ & =4^4+b^4+6(a b)^2+4 a b \times\left[(a+b)^2-2 a b\right] \\ & =a^4+4^4+6 \times 2^2+4 \times 2 \times\left[5^2-2 \times 2\right] \\ & =a^4+b^4+24+8 \times[25-4] \\ & =a^4+b^4+24+8 \times 21 \\ = & a^4+b^4+192\end{aligned}$[/tex]

Since a+b=5 ,then we get this equation :

5⁴ = a⁴ + b⁴ + 168

then

a⁴ + b⁴ = 5⁴ - 192

            = 625 - 192

            = 433

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