Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Simplify the following:
$$(\sqrt{3}-\sqrt{2})^{2}$$ is equal to $$5-2\sqrt{6}$$.
If true then enter $$1$$ and if false then enter $$0$$.

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$-2$$

Solution

We know, $$(a-b)^{2}$$ = $$a^{2} + b^{2} - 2{a}{b}$$.

Then, $$(\sqrt{3}-\sqrt{2})^{2}$$
= $$(\sqrt{3})^{2} + (\sqrt{2})^{2} - 2\times\sqrt{3}\times\sqrt{2}$$
= $$3 + 2 - 2\sqrt{6} $$
= $$5 - 2\sqrt{6}$$.

Hence, the statement is correct.
Therefore, option $$A$$ is correct.
Question 2
1 / -0

Use the identity $$(x + a) (x + b) = x^2 + (a + b) x + ab$$ to find the given product:
$$(2x + 5y)$$$$(2x + 3y)$$.

A
$$ 4x^2 + 24xy - 15y^2$$

B
$$ 4x^2 + 16xy + 25y^2$$

C
$$ 4x^2 + 16xy + 15y^2$$

D
$$ x^2 + 16xy + 15y^2$$

Solution

Given, $$(2x + 5y) (2x+3y)$$.

We know, $$(x + a) (x + b) = x^2 + (a + b) x + ab$$.

On comparing, we get,
$$x=2x ,a=5y ,b=3y$$.

Thus,
$$(2x + 5y) (2x+3y)$$
$$=(2x)^2+(5y+3y)2x+(5y)\times(3y)$$
$$=4x^2+(8y)2x+15y^2$$
$$=4x^2+16xy+15y^2$$.

Hence, option $$C$$ is correct.
Question 3
1 / -0

Use the identity $$(x + a) (x + b) = x^2 + (a + b) x + ab$$ to find the given product:
$$(xyz +4) (xyz +2)$$.

A
$$ x^2 y^2 z^2+ 6xyz $$

B
$$ x^2 y^2 z^2 + 8$$

C
$$ x^2 y^2 z^2 + 6xyz + 8$$

D
$$ x^2 y^2 z^2 - 6xyz + 8$$

Solution

Given, $$(xyz+ 4) (xyz+2)$$.

We know, $$(x + a) (x + b) = x^2 + (a + b) x + ab$$.

On comparing, we get,
$$x=xyz ,a=4 ,b=2$$.

Thus,
$$(xyz+4) (xyz+2)$$
$$=(xyz)^2+(4+2)xyz+4\times 2$$
$$=x^2y^2z^2+6xyz+8$$.

Therefore, option $$C$$ is correct.
Question 4
1 / -0

Use the identity $$(x + a) (x + b) = x^2 + (a + b) x + ab$$ to find the given product:
$$(4x-5)$$$$ (4x-1)$$.

A
$$16x^2-24x + 5$$

B
$$16x^2+ 24x + 1$$

C
$$16x^2+ 20x + 4$$

D
$$16x^2+ 20x + 5$$

Solution

Given, $$(4x- 5) (4x -1)$$.

We know, $$(x + a) (x + b) = x^2 + (a + b) x + ab$$.

On comparing, we get,
$$x=4x ,a=-5 ,b=-1$$.

Thus,
$$=(4x- 5) (4x -1)$$
$$=(4x)^2+[(-5)+(-1)]4x+(-5)\times(- 1)$$
$$=16x^2+(-6)4x+(5)$$
$$=16x^2-24x+5$$.

Hence, option $$A$$ is correct.
Question 5
1 / -0

Use the identity $$(x + a) (x + b) = x^2 + (a + b) x + ab$$ to find the given product:
$$(4x + 5)$$$$(4x-1)$$.

A
$$16x^2 + 16x - 5$$

B
$$16x^2 - 16x - 5$$

C
$$16x^2 + 16x + 5$$

D
$$16x^2 + 12x - 5$$

Solution

Given, $$(4x + 5) (4x- 1)$$.

We know, $$(x + a) (x + b) = x^2 + (a + b) x + ab$$.

On comparing, we get,
$$x=4x ,a=5 ,b=-1$$.

Thus,
$$=(4x+ 5) (4x -1)$$
$$=(4x)^2+[(5)+(-1)]4x+(5)\times(- 1)$$
$$=16x^2+(4)4x+(-5)$$
$$=16x^2+16x-5$$.

Hence, option $$A$$ is correct.
Question 6
1 / -0

Simplify :
$$(4m + 5n)^2 + (5m + 4n)^2$$.

A
$$41m^2 + 80mn + 41n^2$$

B
$$4m^2 - 54mn + 41n^2$$

C
$$41m + 80mn + 4n^2$$

D
$$41m^2 + 10mn + 41n^2$$

Solution

Given, $$(4m + 5n)^2 + (5m + 4n)^2$$.

We know, $$(a+b)^2=a^2+2ab+b^2$$.

Then,
$$(4m + 5n)^2 + (5m + 4n)^2$$
$$=[(4m)^2+2(4m)(5n)+(5n)^2]+[(5m)^2+2(5m)(4n)+(4n)^2]$$
$$=16m^2+40mn+25n^2+25m^2+40mn+16n^2$$
$$=41m^2+80mn+41n^2$$.

Therefore, option $$A$$ is correct.
Question 7
1 / -0

Simplify:
$$(m^2 - n^2m)^2 + 2m^3n^2$$.

A
$$m^4 - n^4m^2$$

B
$$m^4 + n^4m^2$$

C
$$m^4 + n^4$$

D
$$m^4 + n^4m^2 + 2m^3n^2$$

Solution

Given, $$(m^2 - n^2m)^2 + 2m^3n^2$$.

We know, $$(a-b)^2=a^2-2ab+b^2$$.

Then,
$$(m^2 - n^2m)^2 + 2m^3n^2$$
$$=(m^2)^2-2(m^2)(n^2m)+(n^2m)^2+2m^3n^2$$
$$=m^4-2m^3n^2+n^4m^2+2m^3n^2$$
$$=m^4+n^4m^2$$.

Therefore, option $$B$$ is correct.
Question 8
1 / -0

The product of $$4a^{2}, -6b^{2}$$ and $$3a^{2}b^{2}$$ is

A
$$a^{2}b^{2}$$

B
$$13a^{4}b^{4}$$

C
$$-72a^{4}b^{4}$$

D
$$a^{4}b^{4}$$
Question 9
1 / -0

Simplify:
$$(ab + bc)^2 - 2ab^2c$$.

A
$$ab + bc$$

B
$$a^2b^2 - b^2c^2$$

C
$$a^2b^2 + b^2c^2$$

D
$$a^2b + c^2b$$

Solution

Given, $$(ab + bc)^2 - 2ab^2c$$.

We know,
$$(a+b)^2=a^2+2ab+b^2$$.

Then,
$$(ab + bc)^2 - 2ab^2c$$
$$=(ab)^2+2(ab)(bc)+(bc)^2-2ab^2c$$
$$=a^2b^2+2ab^2c+b^2c^2-2ab^2c$$
$$=a^2b^2+b^2c^2$$.

Therefore, option $$C$$ is correct.
Question 10
1 / -0

Use the identity $$(x + a) (x + b) = x^2 + (a + b) x + ab$$ to find the given product:
$$(2a^2 + 9)$$$$(2a^2 + 5)$$.

A
$$ 4a^4 + 28a^2 - 5$$

B
$$ 4a^4 + 28a^2 + 45$$

C
$$ 4a^3 + 28a^2 + 45$$

D
$$ a^4 + 28a^2 + 45$$

Solution

Given, $$(2a^2 + 9) (2a^2+5)$$.

We know, $$(x + a) (x + b) = x^2 + (a + b) x + ab$$.

On comparing, we get,
$$x=2a^2 ,a=9 ,b=5$$.

Thus,
$$(2a^2 + 9) (2a^2+5)$$
$$=(2a^2)^2+(9+5)2a^2+(9)\times(5)$$
$$=4a^2+(14)2a^2+45$$
$$=4a^4+28a^2+45$$.

Therefore, option $$B$$ is correct.

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Algebraic Expressions and Identities Test - 23

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Algebraic Expressions and Identities Test - 23

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Question 1

$$1$$

$$0$$

$$-1$$

$$-2$$

Solution

Question 2

$$ 4x^2 + 24xy - 15y^2$$

$$ 4x^2 + 16xy + 25y^2$$

$$ 4x^2 + 16xy + 15y^2$$

$$ x^2 + 16xy + 15y^2$$

Solution

Question 3

$$ x^2 y^2 z^2+ 6xyz $$

$$ x^2 y^2 z^2 + 8$$

$$ x^2 y^2 z^2 + 6xyz + 8$$

$$ x^2 y^2 z^2 - 6xyz + 8$$

Solution

Question 4

$$16x^2-24x + 5$$

$$16x^2+ 24x + 1$$

$$16x^2+ 20x + 4$$

$$16x^2+ 20x + 5$$

Solution

Question 5

$$16x^2 + 16x - 5$$

$$16x^2 - 16x - 5$$

$$16x^2 + 16x + 5$$

$$16x^2 + 12x - 5$$

Solution

Question 6

$$41m^2 + 80mn + 41n^2$$

$$4m^2 - 54mn + 41n^2$$

$$41m + 80mn + 4n^2$$

$$41m^2 + 10mn + 41n^2$$

Solution

Question 7

$$m^4 - n^4m^2$$

$$m^4 + n^4m^2$$

$$m^4 + n^4$$

$$m^4 + n^4m^2 + 2m^3n^2$$

Solution

Question 8

$$a^{2}b^{2}$$

$$13a^{4}b^{4}$$

$$-72a^{4}b^{4}$$

$$a^{4}b^{4}$$

Question 9

$$ab + bc$$

$$a^2b^2 - b^2c^2$$

$$a^2b^2 + b^2c^2$$

$$a^2b + c^2b$$

Solution

Question 10

$$ 4a^4 + 28a^2 - 5$$

$$ 4a^4 + 28a^2 + 45$$

$$ 4a^3 + 28a^2 + 45$$

$$ a^4 + 28a^2 + 45$$

Solution

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