Question

solve the problem given in the attachment in notebook ​

Answer 1

$1.) {x}^{2} + 3x + 1 = (x - 2) {}^{2} \\ = > {x}^{2} + 3x + 1 = x {}^{2} - 4x - 4 \\ = > x {}^{2} - {x}^{2} + 3x + 4x + 1 + 4 = 0 \\ = > 7x + 5 = 0 \\ 2.)(x + 2) {}^{3} = 2x( {x}^{2} - 1) \\ = > {x}^{3} + 8 + 6 {x}^{2} + 12x = 2x {}^{2} - 2 \\ = > {x}^{3} + {6x}^{2} + 12x + 8 - {2x}^{3} + 2x \\ = > -{x}^{3} + {6x}^{2} + 14x + 8 = 0 \\ 3.) {x}^{2} - 2x = ( - 2)(3 - x) \\ = > {x}^{2} - 2x = - 6 + 2x \\ = > {x}^{2} - 2x - 2x + 6 = 0 \\ = > {x}^{2} - 4x + 6 = 0 \\ 4.)(x + 1) {}^{2} = 2(x - 3) \\ = > {x}^{2} + 2x + 1 = 2x - 6 \\ = > {x}^{2} + 2x - 2x + 1 + 6 = 0 \\ = > {x}^{2} + 7 = 0$

Hence the question no 3 and 4 are quadratic equation.. because these are in the form of x^2+bx+c=0(where x is not equal to zero)..and question no 1 and 2 are not quadratic equation..

$\textbf{\pink{\large{\underline{Hope it helps you... thanks}}}}$

Answer 2

refer to the attachment!!!