Which of the following are quadratic equations?(i)x2+6x−4=0 (ii)√3x2−2x+12=0(iii)x2+1x2=5(iv)x−3x=x2(v)2x2−√3x+9=0(vi)x2−2x−√x−5=0(vii)3x2−5x+9=x2−7x+3(viii)x+1x=1(ix)x2−3x=0(x)(x+1x)2=3(x+1x)+4(xi)(2x+1)(3x+2)=6(x−1)(x−2)(xii)x+1x=x2,x≠0(xiii)16x2−3=(2x+5)(5x−3)(xiv)(x+2)3=x3−4(xv)x(x+1)+8=(x÷2)(x−2)

Which of the following are quadratic equations?(i)x2+6x−4=0 (ii)√3

1.	Which of the following are quadratic equations?(i)x2+6x−4=0 (ii)√3x2−2x+12=0(iii)x2+1x2=5(iv)x−3x=x2(v)2x2−√3x+9=0(vi)x2−2x−√x−5=0(vii)3x2−5x+9=x2−7x+3(viii)x+1x=1(ix)x2−3x=0(x)(x+1x)2=3(x+1x)+4(xi)(2x+1)(3x+2)=6(x−1)(x−2)(xii)x+1x=x2,x≠0(xiii)16x2−3=(2x+5)(5x−3)(xiv)(x+2)3=x3−4(xv)x(x+1)+8=(x÷2)(x−2)
Answer» Which of the following are quadratic equations? $(i) x^{2} + 6 x - 4 = 0$ ( $i i) \sqrt{3 x^{2}} - 2 x + \frac{1}{2} = 0$ $(i i i) x^{2} + \frac{1}{x^{2}} = 5$ $(i v) x - \frac{3}{x} = x^{2}$ $(v) 2 x^{2} - \sqrt{3 x} + 9 = 0$ $(v i) x^{2} - 2 x - \sqrt{x} - 5 = 0$ $(v i i) 3 x^{2} - 5 x + 9 = x^{2} - 7 x + 3$ $(v i i i) x + \frac{1}{x} = 1$ $(i x) x^{2} - 3 x = 0$ $(x) {(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4$ $(x i) (2 x + 1) (3 x + 2) = 6 (x - 1) (x - 2)$ $(x i i) x + \frac{1}{x} = x^{2}, x \neq 0$ $(x i i i) 16 x^{2} - 3 = (2 x + 5) (5 x - 3)$ $(x i v) (x + 2)^{3} = x^{3} - 4$ $(x v) x (x + 1) + 8 = (x \div 2) (x - 2)$

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