4x²+12x+9
b²-4ac
(12)²-4×4×9
144-144
0
[tex]\large\underline{\sf{Solution-4}}[/tex]
Given algebraic expression is
[tex]\sf\: {5x}^{23} - {3x}^{32} + 2x - 12 \\ [/tex]
On substituting x = 1, we get
[tex]\sf\: = \: {5(1)}^{23} - {3(1)}^{32} + 2(1) - 12 \\ [/tex]
[tex]\sf\: = \: 5 - 3 + 2 - 12 \\ [/tex]
[tex]\sf\: = \: 7 - 15 \\ [/tex]
[tex]\sf\: = \: - 8\\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:(d) \: \: None\:of\:these \: } \\ [/tex]
[tex]\large\underline{\sf{Solution-5}}[/tex]
Given expression is
[tex]\sf\: \dfrac{ {38}^{2} - {22}^{2} }{16} \\ [/tex]
[tex]\sf\: = \: \dfrac{(38 - 22)(38 + 22)}{16} \: \: \: \{ \because \: {x}^{2} - {y}^{2} = (x - y)(x + y) \} \\ [/tex]
[tex]\sf\: = \: \dfrac{16 \times 60}{16} \\ [/tex]
[tex]\sf\: = \:60\\ [/tex]
[tex]\implies\sf\:\boxed{\bf\:\dfrac{ {38}^{2} - {22}^{2} }{16} = 60 \: }\\ [/tex]
[tex]\large\underline{\sf{Solution-6}}[/tex]
We know, Pythagorean triplets are represented as
[tex]\sf\: 2m, {m}^{2} - 1, {m}^{2} + 1 \\ [/tex]
Now, it is given that one of the Pythagorean triplet is 6
Let assume that 2m = 6
[tex]\implies\sf\:m = 3 \\ [/tex]
So,
[tex]\begin{gathered}\begin{gathered}\bf\: Pythagorean \: triplet \: are-\begin{cases} &\sf{2m = 6} \\ \\ &\sf{ {m}^{2} - 1 = {3}^{2} - 1 = 8 } \\ \\ &\sf{ {m}^{2} + 1 = {3}^{2} + 1 = 10 } \end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\large\underline{\sf{Solution-7}}[/tex]
[tex]\sf\: {4x}^{2} + 12x + 9 \\ [/tex]
[tex]\sf\: = \: {(2x)}^{2} + 2(2x)(3) + {(3)}^{2} \\ [/tex]
We know,
[tex]\boxed{\sf\: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2} \: } \\ [/tex]
So, Using this algebraic identity, we get
[tex]\sf\: = \: {(2x + 3)}^{2} \\ [/tex]
[tex]\sf\: = \: (2x + 3)(2x + 3) \\ [/tex]
[tex]\implies\sf\:\boxed{\bf\: {4x}^{2} + 12x + 9 = \: (2x + 3)(2x + 3) \: }\\ [/tex]
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Answers & Comments
4x²+12x+9
b²-4ac
(12)²-4×4×9
144-144
0
Verified answer
[tex]\large\underline{\sf{Solution-4}}[/tex]
Given algebraic expression is
[tex]\sf\: {5x}^{23} - {3x}^{32} + 2x - 12 \\ [/tex]
On substituting x = 1, we get
[tex]\sf\: = \: {5(1)}^{23} - {3(1)}^{32} + 2(1) - 12 \\ [/tex]
[tex]\sf\: = \: 5 - 3 + 2 - 12 \\ [/tex]
[tex]\sf\: = \: 7 - 15 \\ [/tex]
[tex]\sf\: = \: - 8\\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:(d) \: \: None\:of\:these \: } \\ [/tex]
[tex]\large\underline{\sf{Solution-5}}[/tex]
Given expression is
[tex]\sf\: \dfrac{ {38}^{2} - {22}^{2} }{16} \\ [/tex]
[tex]\sf\: = \: \dfrac{(38 - 22)(38 + 22)}{16} \: \: \: \{ \because \: {x}^{2} - {y}^{2} = (x - y)(x + y) \} \\ [/tex]
[tex]\sf\: = \: \dfrac{16 \times 60}{16} \\ [/tex]
[tex]\sf\: = \:60\\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:\dfrac{ {38}^{2} - {22}^{2} }{16} = 60 \: }\\ [/tex]
[tex]\large\underline{\sf{Solution-6}}[/tex]
We know, Pythagorean triplets are represented as
[tex]\sf\: 2m, {m}^{2} - 1, {m}^{2} + 1 \\ [/tex]
Now, it is given that one of the Pythagorean triplet is 6
Let assume that 2m = 6
[tex]\implies\sf\:m = 3 \\ [/tex]
So,
[tex]\begin{gathered}\begin{gathered}\bf\: Pythagorean \: triplet \: are-\begin{cases} &\sf{2m = 6} \\ \\ &\sf{ {m}^{2} - 1 = {3}^{2} - 1 = 8 } \\ \\ &\sf{ {m}^{2} + 1 = {3}^{2} + 1 = 10 } \end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\large\underline{\sf{Solution-7}}[/tex]
Given algebraic expression is
[tex]\sf\: {4x}^{2} + 12x + 9 \\ [/tex]
[tex]\sf\: = \: {(2x)}^{2} + 2(2x)(3) + {(3)}^{2} \\ [/tex]
We know,
[tex]\boxed{\sf\: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2} \: } \\ [/tex]
So, Using this algebraic identity, we get
[tex]\sf\: = \: {(2x + 3)}^{2} \\ [/tex]
[tex]\sf\: = \: (2x + 3)(2x + 3) \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: {4x}^{2} + 12x + 9 = \: (2x + 3)(2x + 3) \: }\\ [/tex]