Question

In each of the following, determine whether the given values are solutions of the given equation or not:

(i) \(x^{2}-3x+2=0,x=2,x=1\) 

(ii) \(x^{2}+x+1=0,x=0,x=1\)

(iii) \(x^{2}-3\sqrt{3}+6=0,x=\sqrt{3},x=-2\sqrt{3}\)

(iv) \(x+\frac{1}{x}=\frac{13}{6},x=\frac{5}{6},x=\frac{4}{3}\)

(v) \(2x^{2}-x+9=x^{2}+4x+3,x=2,x=3\)

(vi) \(x^{2}-\sqrt{2}x-4=0,=-\sqrt{2},x=-2\sqrt{2}\)

(vii) \(a^{2}x^{2}-3abx+2b^{2}=0,x=a/b,x=b/a\)

Answer 1

(i) \(x^{2}-3x+2=0,x=2,x=1\) 

For x = 2,

2² – 3 × 2 + 2 = 0

⇒ 0 = 0

Thus, x = 2 is a solution.

For, x = 1 

1² – 3 × 1 + 2 = 0 

⇒ 0 = 0 

Thus, x = 1 is a solution.

(ii) \(x^{2}+x+1=0,x=0,x=1\)

For x = 0, 

⇒ 0 + 0 + 1 = 0 

⇒ 1 = 0 which is not true thus x = 0 is not a solution 

For x = 1, 

⇒ 1 + 1 + 1 = 0 

⇒ 3 = 0 which is not true thus x = 1 is not a solution

(iii) \(x^{2}-3\sqrt{3}+6=0,x=\sqrt{3},x=-2\sqrt{3}\)

For x= √3 

⇒ 3 – 3√3 × √3 + 6 = 0 

⇒ 3 – 9 + 6 = 0 

⇒ 0 = 0 

Thus, x = √3 is a solution 

For x = -2√3 

⇒ (-2√3)² – 3√3 × -2√3 + 6 = 0 

⇒ 4 × 3 + 18 + 6 = 0 

⇒ 36 = 0 which is not true, thus x = -2√3 is not a solution

(iv) \(x+\frac{1}{x}=\frac{13}{6},x=\frac{5}{6},x=\frac{4}{3}\)

For x = 5/6

\(\Rightarrow \frac{5}{6}+\frac{6}{5}=\frac{13}{6}\)

\(\Rightarrow \frac{61}{30}=\frac{13}{6}\)

⇒ 61 = 65 which is not true, thus x = 5/6 is not a solution

For x = 4/3

\(\Rightarrow\frac{4}{3}+\frac{3}{4}=\frac{13}{6}\)

⇒ 25/12 = 13/6 

⇒ 25 = 26 which is not true, thus x = 4/3 is not a solution

(v) \(2x^{2}-x+9=x^{2}+4x+3,x=2,x=3\)

For x = 2, 

⇒ 2 × 4 – 2 + 9 = 4 + 4 × 2 + 3 

⇒ 15 = 15, thus x = 2 is a solution. 

For x = 3 

⇒ 2 × 9 – 3 + 9 = 9 + 4 × 3 + 3 

⇒ 24 = 24, thus x = 3 is also a solution

(vi) \(x^{2}-\sqrt{2}x-4=0,=-\sqrt{2},x=-2\sqrt{2}\)

For x = -√2, 

⇒ 2 - √2 × -√2 – 4 = 0 

⇒ 2 + 2 – 4 = 0 

⇒ 0 = 0 

Thus, x = -√2 is a solution 

For x = -2√2 

⇒ 4 × 2 - √2 × -2√2 – 4 = 0 

⇒ 8 + 8 – 4 = 0 

⇒ 12 = 0 which is not true, thus x = -2√2 is not a solution

(vii) \(a^{2}x^{2}-3abx+2b^{2}=0,x=a/b,x=b/a\)

For, x = a/b

\(\Rightarrow a^{2}\times\frac{a^{2}}{b^{2}}-3ab\times\frac{a}{b}+2\times b^{2}=0\)

⇒ a⁴/b² – 3a² + 2b² = 0 which is not true, thus x = a/b is not a solution

For x = b/a

\(\Rightarrow a^{2}\times\frac{a^{2}}{b^{2}}-3ab\times\frac{b}{a}+2 b^{2}=0\)

⇒ b² – 3b² + 2b² = 0 

⇒ 0 = 0 , 

thus x = b/a is a solution