How do you evaluate #\frac{y^{2}-3}{2}+\frac{y}{4}=1#?

1 Answer
Pallu
Nov 17, 2017

# therefore y= -2.5 or y=2#

Explanation:

#\frac{y^{2}-3}{2}+\frac{y}{4}=1#

#=> \frac{y^{2}-3}{2} (2/2)+\frac{y}{4}=1#

#=> \frac{2y^{2}-6}{4} +\frac{y}{4}=1#

#=> \frac{2y^{2}-6 +y}{4} =1#

#=> (2y^{2}-6 +y) =1 xx 4#

#=> 2y^2 +y-6 -4 =0#

#=> 2y^2 +y-10 =0#

To solve this quadratic equation, find two such numbers whose sum is equal to the coefficient of middle term (i.e. 1) and product is equal to the product of coefficients of first and last term (i. e., #2 xx-10= -20)# .

Two such numbers are 5 and -4. So we rewrite the equation as:

#=> 2y^2 -4y +5y -10 =0#

#=> 2y (y - 2) +5(y -2) =0#

#=> (2y +5)(y-2)=0#

#=> (2y+ 5) =0 or (y-2)=0#

# therefore y= -5/2= -2.5 or y=2#

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