Question

if 2 is a root of the quadratic equation 3x^2+ px -8 =0 And the quadratic equation 4x^2-2px+k =0 has equal roots , find the value of k​

Answer 1

Step-by-step explanation:

Let p(x)= 3x^2 +px -8= 0

Since , 2 is the root of the equation

→p(2)= 0

→3(2)^2+2p-8= 0

→12-8+2p=0

→4+2p= 0

→2p= -4

→p=-4/2

→p=-2 .

Now , put p= -2 in 4x^2 -2px+k= 0

→4x^2 +4x+k= 0

Since , the roots are equal

→D= 0.

→4^2-16k= 0

→16-16k=0

→16k= 16

→k=16/16=1

Hence , required value if k=1

Answer 2

Answer:

first we have to find the value of "p"

given that 2 is a root of the quadratic equation

$3x {}^{2} + px - 8 = 0$

so the value of P is

$3(2) {}^{2} + p(2) - 8 = 0$

$3(4) + 2p - 8 = 0$

$12 + 2p - 8 = 0$

$4 + 2p = 0$

$2p = - 4$

$p = - 2$

putting the value of 'p' in

$4x {}^{2} - 2px + k$

$4x {}^{2} - 2(2)x + k$

$4x {}^{2} - 4x + k$

Given that the quadratic equation has equal roots ,then the condition is

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

Value of

$\large{b=-4}$

$\large{a=4}$

$\large{c=k}$

The value of K is

$( - 4) {}^{2} - 4(4)(k) = 0$

$16 - 16k = 0$

$- 16k = - 16$

$k = 1$