Question

for what values of k are the roots of the quadratic equation 3x²+ 2kx+27=0 real and equal?​

Answer 1

Answer:

Let's solve for k.

3x2+2kx+27=0

Step 1: Add -3x^2 to both sides.

2kx+3x2+27+−3x2=0+−3x2

2kx+27=−3x2

Step 2: Add -27 to both sides.

2kx+27+−27=−3x2+−27

2kx=−3x2−27

Step 3: Divide both sides by 2x.

2kx/2x=−3x2−27/2x

k=−3x2−27/2x

Step-by-step explanation:

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Answer 2

a=3

b=2k ; b^2=4k^2

c=27

b^2-4ac=0

b^2=4ac

4k^2=4(3×27)

divide by 4

k^2=81

k=+_9 (plus or minus 9)

so the equation is

3x²+_18x+27=0

divide by 3

x²+_6x+9=0

so

x²+6x+9=0 x²-6x+9=0

x²+3x+3x+9=0 x²-3x-3x+9=0

x (x+3)+3 (x+3)=0 x (x-3)-3 (x-3)=0

x+3=0 x-3=0

x= -3 x= +3