Question

using a quadratic formula show that the equation x^2-8x+18=0 has no solution​

Answer 1

Step-by-step explanation:

Equating the given equation with

ax^2+bx+c=0

We get , a=1, b=-8 and c=18

Using quadratic formula ,

Discriminant , D = b^2-4ac= (-8)^2-4(1)18=64-72=-18

Since , D is less than 0

→The equation has no roots

#Answer with quality

#BAL

Answer 2

Step-by-step explanation:

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.