Question

using completing the square method, show that the equation x^2-8x+18 =0 has no solution

Answer 1

$x^{2}$-8x+18=0

$x^{2}$-8x=-18

$x^{2}$-8x+$4^{2}$=-18+$4^{2}$

$(x-4)^{2}$= -18+16

x-4=$\sqrt{-2}$

sice under root -2 will give a imaginary number therefore this system of equation has no solution

Answer 2

x² - 8x + 18 = 0

or, x² - 2 × 4 × x + 18 = 0

or, x² - 2 × 4 × x + (16 + 2) = 0

or, x² - 2 × 4 × x + 4² + 2 = 0

or, x² - 2 × 4 × x + 4² = - 2

here, you should use algebraic Identity , a² - 2ab + b² = 0

so, x² - 2 × 4 × x + 4² = (x - 4)²

e.g., (x - 4)² = -2

but we know, square of any real number can't be negative .

so, (x - 4)² ≠ -2

hence, x² - 8x + 18 = 0 has no real solution.