Answers

2015-07-11T18:39:20+08:00
2x^2+9x+10=0

Ans:
 Guide: Product of Binomial⇒ (a+b)^2 = a^2+2ab+b^2 

2x^2+9x+10=0  ⇒It is easy to complete the square of an equation if the                                     numerical coefficient of the 1st term is 1. So divide both                                   sides by 2.
x^2+(9/2)x+5=0 ⇒ transpose 5 to the right side so it will become -5.
x^2+(9/2)x=-5 ⇒from the left side of this equation we can use the guide above.                         So our value of a=1 set aside the variable, when we square the                         1 the ans. is always 1. Let's come to (2ab): Our value of 2ab=                           (9/2) since we already have the value of a as 1 divide (9/2) by                           2 to eliminate 2 in 2ab and we can find the value of b.
                        Our b=(9/4)
x^2+(9/2)x+(9/4)^2 = -5+(9/4)^2 ⇒we add both sides of the equation by b^2.
                                                b=(9/4) 
x^2+(9/2)x+(9/4)^2 = -5+(81/16)
[x+(9/4)]^2 = (1/16) ⇒use the guide: the left side. And square both sides of the                                 equation to eliminate the squared of the left side.
x+(9/4)=√(1/16)
x+(9/4)=+ and - of (1/4) ⇒the square root of 1/16 is the positive and negative                                          1/4
x=+ and - (1/4)-(9/4)
or 
x= +(1/4)-(9/4)=-2
x= -(1/4)-(9/4)=-5/2

To check: Use the Quadratic Formula 


 

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