Which of the following are quadratic equation in x ?
Question:

Which of the following are quadratic equation in x ?

(i) $x^{2}-x+3=0$

(ii) $2 x^{2}+\frac{5}{2} x-\sqrt{3}=0$

(iii) $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$

(iv) $\frac{1}{3} x^{2}+\frac{1}{5} x-2=0$

(v) $x^{2}-3 x-\sqrt{x}+4=0$

(vi) $x-\frac{6}{x}=3$

(vii) $x+\frac{2}{x}=x^{2}$

(viii) $x^{2}-\frac{1}{x^{2}}=5$

(ix) $(x+2)^{3}=x^{3}-8$

(x) $(2 x+3)(3 x+2)=6(x-1)(x-2)$

(xi) $\left(x+\frac{1}{x}\right)^{2}=2\left(x+\frac{1}{x}\right)+3$

 

 

Solution:

i) $\left(x^{2}-x+3\right)$ is a quadratic polynomial.

$\therefore x^{2}-x+3=0$ is a quadratic equation.

ii) $C$ learly, $\left(2 x^{2}+\frac{5}{2} x-\sqrt{3}\right)$ is a quadratic polynomial.

$\therefore 2 x^{2}+\frac{5}{2} x-\sqrt{3}=0$ is a quadratic equation.

iii) Clearly, $\left(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\right)$ is a quadratic polynomial.

$\therefore \sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ is a quadratic equation.

iv) Clearly, $\left(\frac{1}{3} x^{2}+\frac{1}{5} x-2\right)$ is a quadratic polynomial.

$\therefore \frac{1}{3} x^{2}+\frac{1}{5} x-2=0$ is a quadratic equation.

v) $\left(x^{2}-3 x-\sqrt{x}+4\right)$ contains a term with $\sqrt{x}, i . e, x^{\frac{1}{2}}$, where $\frac{1}{2}$ is not a integer. Therefore, it is not a quadratic polynomial.

$\therefore x^{2}-3 x-\sqrt{x}+4=0$ is not a quadratic equation.

vi) $x-\frac{6}{x}=3$

$\Rightarrow x^{2}-6=3 x$

$\Rightarrow x^{2}-3 x-6=0$

$\left(x^{2}-3 x-6\right)$ is a quadratic polynomial; therefore, the given

equation is quadratic.

vii) $x+\frac{2}{x}=x^{2}$

$\Rightarrow x^{2}+2=x^{3}$

$\Rightarrow x^{3}-x^{2}-2=0$

$\left(x^{3}-x^{2}-2\right)$ is not a quadratic polynomial.

$\therefore x^{3}-x^{2}-2=0$ is not a quadratic equation.

viii) $x^{2}-\frac{1}{x^{2}}=5$

$\Rightarrow x^{4}-1=5 x^{2}$

$\Rightarrow x^{4}-5 x^{2}-1=0$

$\left(x^{4}-5 x^{2}-1\right)$ is a polynomial with degree 4 .

$\therefore x^{4}-5 x^{2}-1=0$ is not a quadratic equation.

(ix) $(x+2)^{3}=x^{3}-8$

$\Rightarrow x^{3}+6 x^{2}+12 x+8=x^{3}-8$

$\Rightarrow 6 x^{2}+12 x+16=0$

This is of the form axbx + c = 0.
Hence, the given equation is a quadratic equation.

(x) $(2 x+3)(3 x+2)=6(x-1)(x-2)$

$\Rightarrow 6 x^{2}+4 x+9 x+6=6\left(x^{2}-3 x+2\right)$

$\Rightarrow 6 x^{2}+13 x+6=6 x^{2}-18 x+12$

$\Rightarrow 31 x-6=0$

This is not of the form axbx + c = 0.
Hence, the given equation is not a quadratic equation.

(xi) $\left(x+\frac{1}{x}\right)^{2}=2\left(x+\frac{1}{x}\right)+3$

$\Rightarrow\left(\frac{x^{2}+1}{x}\right)^{2}=2\left(\frac{x^{2}+1}{x}\right)+3$

$\Rightarrow\left(x^{2}+1\right)^{2}=2 x\left(x^{2}+1\right)+3 x^{2}$

$\Rightarrow x^{4}+2 x^{2}+1=2 x^{3}+2 x+3 x^{2}$

$\Rightarrow x^{4}-2 x^{3}-x^{2}-2 x+1=0$

This is not of the form axbx + c = 0.
Hence, the given equation is not a quadratic equation.

 

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