How do you find all zeros of the function #f(x)=4x^2+29x+30#?

1 Answer

Answer:

#x_1=-5/4# and #x_2=-6#

Explanation:

From the given equation #f(x) = 4x^2+29x+30#

Set #f(x)=0#

# 4x^2+29x+30=0#

Try factoring

#(4x+5)(x+6)=0#

Set both factors to zero to solve for the roots

#4x+5=0#

#4x=-5#

#x=-5/4# this is a zero

the other one

#x+6=0#

#x=-6#
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We can use another method. Using the Quadratic Formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

From # 4x^2+29x+30=0#, let #a=4# , #b=29#, #c=30#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-29+-sqrt(29^2-4(4)(30)))/(2(4))#

#x=(-29+-sqrt(841-480))/(8)#

#x=(-29+-sqrt(361))/(8)#

#x=(-29+-19)/(8)#

#x_1=(-10)/(8)=-5/4#

#x_2=(-48)/(8)=-6#

God bless....I hope the explanation is useful