Question

can u please answer this

Answer 1

Question 1 :

• Simplify $\sf\:a(a^{2}-a+1)+5a$ and find it's value for a = -2

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Solution :

$\sf\:a(a^{2}-a+1)+5a$

Plugging the value of a as (-2) -

$\sf\dashrightarrow\:(-2)[(-2)^{2}-(-2)+1]+5(-2)$

$\sf\dashrightarrow\:(-2)[4+2+1]+(-10)$

$\sf\dashrightarrow\:(-2)[6+1]-10$

$\sf\dashrightarrow\:(-2)[7]-10$

$\sf\dashrightarrow\:(-2) \times 7-10$

$\sf\dashrightarrow\:-14-10$

$\small{\underline{\boxed{\mathrm\purple{\dashrightarrow(-24)}}}}$ ★

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Question 2 :

• Evaluate using suitable identity : 101 × 103

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Solution :

$\sf\:101 \times 103$

• 101 can be written as ( 100+1 )

• 103 can be written as ( 100+3 )

$\sf\leadsto\:(100+1) \times (100+3)$

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Using an identity :

• $\tt\:(x+a)(x+b) = x^{2}+(a+b) x+ab$

Here :

• x = 100 , a = 1 and b = 3

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$\sf\leadsto\:(100)^{2}+(1+3)100+1 \times 3$

$\sf\leadsto\:10000+4 \times 100+1 \times 3$

$\sf\leadsto\:10000+400+3$

$\sf\leadsto\:10400+3$

$\small{\underline{\boxed{\mathrm\purple{\leadsto\:10403}}}}$ ★

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Question 3 :

• If p + q = 12 and pq = 22 , then find $\sf\:p^{2}+q^{2}$

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Solution :

We know that -

• $\sf\:(a+b) ^{2}=a^{2}+2ab+b^{2}$

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Using this identity in this question -

$\sf\twoheadrightarrow\:(p+q) ^{2}=p^{2}+2pq+q^{2}$

Substituting :

• p + q = 12 and pq = 22

$\sf\twoheadrightarrow\:(12) ^{2}=p^{2}+2 \times 22+q^{2}$

$\sf\twoheadrightarrow\:144=p^{2}+44+q^{2}$

Transposing +44 to LHS it becomes -44

$\sf\twoheadrightarrow\:144-44=p^{2}+q^{2}$

$\sf\twoheadrightarrow\:100=p^{2}+q^{2}$

$\small{\underline{\boxed{\mathrm\purple{\twoheadrightarrow\:oR~p^{2}+q^{2}=100}}}}$ ★

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Therefore , the final answer :

$\sf\color{cyan}{⋆}$ Question 1 answer = $\bf\color{black}{\:(-24)}$

$\sf\color{cyan}{⋆}$ Question 2 answer = $\bf\color{black}{\:10403}$

$\sf\color{cyan}{⋆}$ Question 3 answer => $\bf\color{black}{\:p^{2}+q^{2}=100}$

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