Question

show that the equation ( x-a) (x-b) = h^2 has real roots​

Answer 1

Answer:

Problem : (x - a)(x - b) = h²

=> x² - (a + b)x + ab - h² = 0 - - - (i)

For the Quadratic Equation, ax² + bx + c = 0 ;

x = [-b ± √(b² - 4ac) ] / 2a

and for real roots, discriminant D = b² - 4ac ≥ 0

therefore, for the above quadratic equation, a = 1, b = - (a + b) and c = ab - h²

=> D = [ - (a + b) ]² - 4 * 1 * (ab - h²)

=> D = (a + b) ² - 4ab + 4h²

=> D = (a - b) ² + 4h²

=> D ≥ 0 { as sum of the square quantites will always be positive }

So, the roots of the Q.E. given will always be real.

Further, if h = 6i, then h² = -36

=> D = (a - b) ² + 4(-36)

=> (a - b) ² - 144

=> (a - b) ² - 12²

=> (a - b - 12)(a - b + 12)

for real roots, (a - b - 12)(a - b + 12) ≥ 0

=> (a - b) = ± 12 OR,

(a - b) < -12 OR (a - b) > 12