Question

Find wheather x²+5√5x-70=0 has real roots.If real roots exists,find them​

Answer 1

Solution

→ x² + 5√5x - 70 = 0

We know that,

• D > 0, real and distinct roots
• D = 0, real and equal roots
• D < 0, imaginary roots

→ D = b² - 4ac

→ D = (5√5)² - 4(1)(- 70)

→ D = 125 + 280

→ D = 405

Hence it has real roots.

→ x = (- b ± √D)/2a

[Shridharacharya Formula]

→ x = (- 5√5 ± √405)/2(1)

→ x = (- 5√5 ± 9√5)/2

Case 1: If 9√5 is positive,

→ x = (- 5√5 + 9√5)/2

→ x = 4√5/2 = 2√5

Case 2: If 9√5 is negative,

→ x = (- 5√5 - 9√5)/2

→ x = - 14√5/2 = - 7√5

Hence roots of x are 2√5 and - 7√5.