How do you solve #(4x+5)^2=35x+29# using the quadratic formula?

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Answer 1 · 2016-08-24T12:41:49

The Solns. are #x_1~=0.55425, x_2~=-1.80425#.

Formula to find the roots #alpha, and beta# of the quadr. eqn.

# ax^2+bx+c+0# is, #alpha, beta = (-b+-sqrt(b^2-4ac))/(2a)#.

Before proceed to solve the given eqn., let us first simplify it :

#(4x+5)^2=35x+29#.

#rArr 16x^2+40x+25-35x-29=0, i.e., 4x^2+5x-4=0#.

So, we have, #a=4, b=5, c=-4#

#:. alpha, beta =(-5+-sqrt(25+64))/8=(-5+-sqrt89)/8#

Taking, #sqrt89~=9.434, we have, the roots

#alpha~=0.55425, beta~=-1.80425#.

Enjoy Maths.!