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

Question

Avery Alexander · Accepted Answer

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.!

Isaiah Anthony · Answer

undefined To solve the equation (4x + 5)^2 = 35x + 29 using the quadratic formula, first expand the left side:

(4x + 5)^2 = (4x + 5)(4x + 5) = 16x^2 + 40x + 25

Now, set the expanded expression equal to 35x + 29:

16x^2 + 40x + 25 = 35x + 29

Rearrange the equation to set it to zero:

16x^2 + 40x - 35x + 25 - 29 = 0

Combine like terms:

16x^2 + 5x - 4 = 0

Now, apply the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 16, b = 5, and c = -4.

Plug these values into the quadratic formula:

x = (-(5) ± √((5)^2 - 4(16)(-4))) / (2(16))

Calculate the discriminant:

b^2 - 4ac = (5)^2 - 4(16)(-4) = 25 + 256 = 281

Now, plug the discriminant into the quadratic formula:

x = (-(5) ± √(281)) / (32)

So, the solutions are:

x = (-5 + √281) / 32 and x = (-5 - √281) / 32