Find the values of k that will make each quadratic a perfect square (1 Viewer)

[jjuunnee]

a) x^2 - 2kx + 16

b) (5k - 1)x^2 - 4x + (2k - 1)

Do I solve these using the discriminant formula?

 

[InteGrand]

a) x^2 - 2kx + 16

b) (5k - 1)x^2 - 4x + (2k - 1)

Do I solve these using the discriminant formula?

Yes. You'll want (and need) the discriminant to be 0.

 

[davidgoes4wce]

 

[InteGrand]

We can also have k = -4.

 

[Green Yoda]

We can also have k = -4.

oh yeah forgot about that, it is squared so it must be +-4

 

[Green Yoda]

How would you do this formally? I just did it by inspection.

 

[InteGrand]

How would you do this formally? I just did it by inspection.

Set the discriminant to 0 and solve for k.

 

[Green Yoda]

Wait my bad when delta=0 it is a perfect square...so just solve for k

 

[Green Yoda]

@OP
a) Δ=4k^2-4*16=0
therefore k^2=16
therefore k=+-4

b)Δ=16-4(5k-1)(2k-1)=0
therefore 16-4(10k^2-7k+1)=0
therefore 16-40k^2+28k-4=0
therefore 40k^2-28k-12=0
therefore 10k^2-7k-3=0
therefore k=1 or -0.3