Question

solve the following quadratic equation by completing square method : x2+2x-5=0

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Answer 1

Answer:

We have,

2x2−5x+3=0

⇒x2−25x+23=0 (Dividing throughough by 2)

⇒x2−25x=−23 (Shifting the constant term on RHS)

⇒x2−2(45)x+(45)2=(45)2−23 (Adding (21Coeff.ofx)2 onboth sides)

⇒(x−45)2=1625−23⇒(x−45)2=161⇒x−45=±41⇒x=45±41

⇒x=45+41=46orx=45−41=44⇒x=23orx=1

Here the roots of the equation are 23 and 1

Answer 2

Step-by-step explanation:

x2+2x-5=0

x2+2x+1-1-5=0

x2+2x+1-6=0

(x+1)2=6

x+1=+-√6

x=√6-1

x=-√6-1