Question

Find k in equation 9x square+6kx+4=0 has equal roots

Answer 1

Given an equation

$\red{9x^{2} + 6kx + 4 = 0}$

And we need to find out the value of K for which the quadratic equation has real and equal roots

So let us find the value of K

We know that for the equal and real roots the descriminent is equal to zero

I,e

$\bold{b^{2} - 4ac = 0}$

So from the above equation we get

$a = 9 \: b = 6k \: and c = 4$

Let us out the values of a,band c in the equation b²-4ac so as to get the value of K

$= > b ^{2} - 4ac = 0 \\ \\ = > (6k)^{2} - 4 \times 9 \times 4 = 0 \\ \\ = >36k ^{2} - 144 = 0 \\ \\ = > 36k ^{2} = 144 \\ \\ = > k ^{2} = \frac{144}{36} \\ \\ = > k ^{2} = 4 \\ \\ = > k = ±\sqrt{4} \\ \\ = > \bold{ \boxed{ \red{k =± 2}}}$

The value of K for which the equation has real and equal roots is 2 ✓

Answer 2

Answer:

If the given equation has equal roots show the discriminant will be zero

i.e.

${(6k)}^{2} - 4 \times 9 \times 4 = 0 \\ \\ 36 {k}^{2} - 144 = 0 \\ \\ 36 {k}^{2} = 144 \\ \\ {k}^{2} = \frac{144}{36} \\ \\ {k}^{2} = 4 \\ \\ k = \pm2$

So,

for k = 2, - 2 given equation will have two equal roots

which are given by

x = -2/3 if k = 2

and

x = 2/3 if k = 2