Question

Find the value of
a) 3p² + 4q² − 5 when p = 3 and q = −2
b) x³ − 3x²y + 2xy² + 8xy + 9 when x = −3 and y = 1.

Answer 1

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Answer 2

$\large \underline \text{Question: -}$

Find the value of

• a) 3p² + 4q² − 5 when p = 3 and q = −2

• b) x³ − 3x²y + 2xy² + 8xy + 9 when x = −3 and y = 1

$\large \underline \text{Solution: -}$

a) 3p² + 4q² − 5 when p = 3 and q = −2

Given that,

• $p = 3 \: \: \text{and} \: \: q = - 2$

We have,

• $3 {p}^{2} + 4 {q}^{2} - 5$

After substituting the given values,

$\implies 3 {p}^{2} + 4 {q}^{2} - 5 \\ \\ \implies 3 {(3)}^{2} + 4 {( - 2)}^{2} - 5 \\ \\ \implies 3(9) + 4(4) - 5 \\ \\ \implies 27 + 16 - 5 \\ \\ \implies 43 - 5 \\ \\ \implies 38$

Therefore,

• 3p² + 4q² − 5 is equals to 38 when p = 3 and q = −2.

.

b) x³ − 3x²y + 2xy² + 8xy + 9 when x = −3 and y = 1

Given that,

• $x = - 3 \: \: \text{and} \: \: y = 1$

We have,

• ${x}^{3} - 3 {x}^{2} y + 2x {y}^{2} + 8xy + 9$

After substituting the given values,

$\implies {x}^{3} - 3 {x}^{2} y + 2x {y}^{2} + 8xy + 9 \\ \\ \implies {( - 3)}^{3} - 3 {( - 3)}^{2} (1) + 2( - 3) {(1)}^{2} + 8( - 3)1 + 9 \\ \\ \implies - 27 - 3(9)1 - 6(1) - 24 + 9 \\ \\ \implies - 27 - 27 - 6 - 24 + 9 \\ \\ \implies - 84 + 9 \\ \\ \implies - 75$

Therefore,

• x³ − 3x²y + 2xy² + 8xy + 9 is equals to - 75 when x = −3 and y = 1.

$\\ \large \underline \text{Required Answer: -}$

• a) 3p² + 4q² − 5 is equals to 38 when p = 3 and q = −2.

• b) x³ − 3x²y + 2xy² + 8xy + 9 is equals to - 75 when x = −3 and y = 1.