MHB Find K Values: Solve Equations w/ One Solution | Help

[Drain Brain]

Find all values of k that ensure that the given equation has
exactly one solution.

1. $4x^2+Kx+25=0$
2.$Kx^2+36x+K=0$

what's the first step to solve these. Help me get started.

Regards

 

[kaliprasad]

Find all values of k that ensure that the given equation has
exactly one solution.

1. $4x^2+Kx+25=0$
2.$Kx^2+36x+K=0$

what's the first step to solve these. Help me get started.

Regards

for this to have exactly one solution we must have both solutions to be same or it has to be a perfect square

that is in $Ax^2 + Bx + C = 0$ we must have $B^2= 4 AC$

in $4x^2+Kx+25=0$ A = 4, B = K and C = 25

 

[Drain Brain]

Oh. I see.

$B^2-4ac=0$

then,

$K^2-4(1)(25)=0$

$K^2=100$

$K=\pm\sqrt{10}$ ---> is this correct?

 

[Evgeny.Makarov]

$B^2-4ac=0$

then,

$K^2-4(1)(25)=0$

Why is 4 multiplied by 1?

 

[kaliprasad]

Oh. I see.

$B^2-4ac=0$

then,

$K^2-4(1)(25)=0$

$K^2=100$

$K=\pm\sqrt{10}$ ---> is this correct?

No , A = 4 but you have taken A = 1

 

[Drain Brain]

Why is 4 multiplied by 1?

I see missed out $a=4$ the answer should be $\pm10\sqrt{2}$

 

[kaliprasad]

I see missed out $a=4$ the answer should be $\pm10\sqrt{2}$

you have to take square root of 400 that is + or - 20 . why square root of 2 ?

 

[Drain Brain]

you have to take square root of 400 that is + or - 20 . why square root of 2 ?

my bad! It seems that I got rusty in noticing little things. Please bear with me.

$K^2-4(4)(25)=0$

$K^2=400$

$K=\pm 20$

 

[Evgeny.Makarov]

$K=\pm 20$

Now it's correct.