Question

PLEASE Solve the following quadratic equations by completing the square
(i) x² + 6x - 7 0
(ii) x² + 3x + 1 0
(in) 2x² + 5x - 3 0
(iv) 4x² + 4bx - (a – 6²) - 0
(v) x² – (√3+1)x+√3 - 0
(vi) Sx + 7 - 3x + 2​

Answer 1

Answer :-

${x}^{2} + 6x - 7 = 0 \\ {x}^{2} + 6x = 7 \\ 3rd \: \: term = ( \frac{1}{2} \times coef. \: of \: x) {}^{2} \\ = {( \frac{1}{2} \times 6) }^{2} \\ = { {3}^{} }^{2} = 9 \\ adding \: third \: term \: on \: both \: side \\ {x}^{2} + 6x + 9 = 7 + 9 \\ {x}^{2} + 6x + 9 = 16 \\ take \: sq.root \: of \: both \: sides \\ (x + 3) {}^{2} = 4 \\ x + 3 = + - 4 \\ x + 3 = 4 \: or \: x + 3 = - 4 \\ x = 1 \: \: or \: \: x = - 7$

I solve 1 example

You can solve like this way

Thanks for the question bro