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