2×55
23
Answers
Explanation:
Given equation is
{\bf :\longmapsto\: { |x - 2| }^{ {10x}^{2} - 1} = { |x - 2| }^{3x}}:⟼∣x−2∣
10x
2
−1
=∣x−2∣
3x
Using hit and trial method,
If we take x = 2, we get
\red{\rm :\longmapsto\: { |2 - 2| }^{ {10x}^{2} - 1} = { |2 - 2| }^{3x}}:⟼∣2−2∣
10x
2
−1
=∣2−2∣
3x
\red{\rm :\longmapsto\: {0 }^{ {10(2)}^{2} - 1} = { 0 }^{3 \times 2}}:⟼0
10(2)
2
−1
=0
3×2
\red{\rm :\longmapsto\: 0 = 0}:⟼0=0
\red{\bf\implies \:x = 2 \: is \: its \: solution}⟹x=2isitssolution
Consider again,
\purple{\bf :\longmapsto\: { |x - 2| }^{ {10x}^{2} - 1} = { |x - 2| }^{3x}}:⟼∣x−2∣
10x
2
−1
=∣x−2∣
3x
Again, If we take x = 3, we get
\purple{\bf :\longmapsto\: { |3 - 2| }^{ {10(3)}^{2} - 1} = { |3 - 2| }^{3 \times 3}}:⟼∣3−2∣
10(3)
2
−1
=∣3−2∣
3×3
\purple{\bf :\longmapsto\: { |1| }^{ 90 - 1} = { |1| }^{9}}:⟼∣1∣
90−1
=∣1∣
9
\purple{\bf :\longmapsto\: { |1| }^{ 89} = { |1| }^{9}}:⟼∣1∣
89
=∣1∣
9
\purple{\bf :\longmapsto\: 1 = 1}:⟼1=1
\purple{\bf\implies \:x = 3 \: is \: its \: solution}⟹x=3isitssolution
Again, Consider
\green{\bf :\longmapsto\: { |x - 2| }^{ {10x}^{2} - 1} = { |x - 2| }^{3x}}:⟼∣x−2∣
10x
2
−1
=∣x−2∣
3x
If we take x = 1, we get
\green{\bf :\\: { |1 - 2| }^{ {10(1)}^{2} - 1} = { |1 - 2| }^{3 \times 1}}:⟼∣1−2∣
10(1)
2
−1
=∣1−2∣
3×1
\green{\bf :\\: { | - 1| }^{ 10 - 1} = { | - 1| }^{3}}:⟼∣−1∣
10−1
=∣−1∣
3
\green{\bf :\\: { | 1| }^{9} = { | 1| }^{3}}:⟼∣1∣
9
=∣1∣
3
\green{\bf :\ = 1 }:⟼1=1
\green{\bf\implies \:x = 1 \: is \: its \: solution}⟹x=1isitssolution
Again Consider
{\bf :\\: { |x - 2| }^{ {10x}^{2} - 1} = { |x - 2| }^{3x}}:⟼∣x−2∣
10x
2
−1
=∣x−2∣
3x
\rm \implies\: {10x}^{2} - 1 = 3x⟹10x
2
−1=3x
\rm \implies\: {10x}^{2} - 3x - 1 = 0⟹10x
2
−3x−1=0
\rm \implies\: {10x}^{2} - 5x + 2x- 1 = 0⟹10x
2
−5x+2x−1=0
\rm :\\:5x(2x - 1) + 1(2x - 1) = 0:⟼5x(2x−1)+1(2x−1)=0
\rm :\\:(2x - 1)(5x + 1) = 0:⟼(2x−1)(5x+1)=0
\bf\implies \:x = \} \: \: or \: \: - \: \}⟹x=
2
1
or−
5
1
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x=2
x=3
x=1
x=−
5
1
x=
2
1