Question

Which of the following are quadratic equation in x ?
(i) $x^2-x+3=0$
(ii) $2x^2+\frac{5}{2}x-\sqrt{3}=0$
(iii) $\sqrt{2}{x}^2+7x+5\sqrt{2}=0$
(iv) $\frac{1}{3}{x}^2+\frac{1}{5}x-2=0$
(v) $x^2-3x-\sqrt{x}+4=0$
(vi) $x-\frac{6}{x}=3$
(vii) $x+\frac{2}{x}=x^2$
(viii) $x^2-\frac{1}{x^2}=5$
(ix) ${\left(x+2\right)}^3=x^3-8$
(x) $\left(2x+3\right)\left(3x+2\right)=6\left(x-1\right)\left(x-2\right)$
(xi) ${\left(x+\frac{1}{x}\right)}^2=2\left(x+\frac{1}{x}\right)+3$

Answer 1

$$\begin{gathered} \mathrm{i})(x^2-x+3)\mathrm{is\; a\; quadratic\; polynomial}. \\ \therefore x^2-x+3=0\mathrm{is\; a\; quadratic\; equation}. \\ \mathrm{ii)}C\mathrm{learly,}(2x^2+\frac{5}{2}x-\sqrt{3})\mathrm{is\; a\; quadratic\; polynomial}. \\ \therefore 2x^2+\frac{5}{2}x-\sqrt{3}=0\mathrm{is\; a\; quadratic\; equation}. \\ \mathrm{iii)}C\mathrm{learly,}(\sqrt{2}{x}^2+7x+5\sqrt{2})\mathrm{is\; a\; quadratic\; polynomial}. \\ \therefore \sqrt{2}{x}^2+7x+5\sqrt{2}=0\mathrm{is\; a\; quadratic\; equation}. \\ \mathrm{iv)}C\mathrm{learly,}(\frac{1}{3}{x}^2+\frac{1}{5}x-2)\mathrm{is\; a\; quadratic\; polynomial}. \\ \therefore \frac{1}{3}{x}^2+\frac{1}{5}x-2=0\mathrm{is\; a\; quadratic\; equation}. \\ \mathrm{v)}(x^2-3x-\sqrt{x}+4)\mathrm{contains\; a\; term}\mathrm{with}\sqrt{x},i\mathrm{.e,}{x}^{\frac{1}{2}},\mathrm{where}\frac{1}{2}is\; not\; a\mathrm{integer}.\mathrm{Therefore},\mathrm{it}\mathrm{is\; not\; a\; quadratic\; polynomial}. \\ \therefore x^2-3x-\sqrt{x}+4=0\mathrm{is\; not\; a\; quadratic\; equation}. \\ \mathrm{vi)}x-\frac{6}{x}=3 \\ \Rightarrow x^2-6=3x \\ \Rightarrow x^2-3x-6=0 \\ \begin{array}{l}(x^2-3x-6)\mathrm{is}\mathrm{a}\mathrm{quadratic}\mathrm{polynomial};\mathrm{therefore},\mathrm{the\; given}\\ \mathrm{equation\; is}\mathrm{quadratic}.\end{array} \\ \mathrm{vii)}x+\frac{2}{x}=x^2 \\ \Rightarrow x^2+2=x^3 \\ \Rightarrow x^3-x^2-2=0 \\ (x^3-x^2-2)\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{quadratic}\mathrm{polynomial}. \\ \therefore x^3-x^2-2=0\mathrm{is\; not\; a\; quadratic\; equation}. \\ \mathrm{viii)}{x}^2-\frac{1}{x^2}=5 \\ \Rightarrow x^4-1=5x^2 \\ \Rightarrow x^4-5x^2-1=0 \\ (x^4-5x^2-1)\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{with}\mathrm{degree}4. \\ \therefore x^4-5x^2-1=0\mathrm{is\; not\; a\; quadratic\; equation}. \end{gathered}$$
(ix) ${\left(x+2\right)}^3=x^3-8$
$$\begin{gathered} \Rightarrow x^3+6x^2+12x+8=x^3-8 \\ \Rightarrow 6x^2+12x+16=0 \end{gathered}$$
This is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $\left(2x+3\right)\left(3x+2\right)=6\left(x-1\right)\left(x-2\right)$
$$\begin{gathered} \Rightarrow 6x^2+4x+9x+6=6\left(x^2-3x+2\right) \\ \Rightarrow 6x^2+13x+6=6x^2-18x+12 \\ \Rightarrow 31x-6=0 \end{gathered}$$
This is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${\left(x+\frac{1}{x}\right)}^2=2\left(x+\frac{1}{x}\right)+3$
$$\begin{gathered} \Rightarrow {\left(\frac{x^2+1}{x}\right)}^2=2\left(\frac{x^2+1}{x}\right)+3 \\ \Rightarrow {\left(x^2+1\right)}^2=2x\left(x^2+1\right)+3x^2 \\ \Rightarrow x^4+2x^2+1=2x^3+2x+3x^2 \\ \Rightarrow x^4-2x^3-x^2-2x+1=0 \end{gathered}$$
This is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.