#y=x^2+7x-5# can be written in the form #y=(x+a)^2+b#. What is the value of #a# and #b#?

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Answer 1 · 2016-09-28T12:19:50

#x^2+7x-5 = (x + 7/2)^2-69/4#

#y = (x+a)^2+b = x^2+2ax+a^2+b#

also

#y = x^2+7x-5#

Making

#{(2a=7),(a^2+b=-5):}#

and solving

#a = 7/2# and #b = -69/4#

we will have

#x^2+7x-5 = (x + 7/2)^2-69/4#