Question

for what values of n the equation 2x ^2 -nx + n =0 has coincident roots​

Answer 1

Step-by-step explanation:

a

2

x

2

+2(a+1)x+4=0

condition for coincident roots b

2

=4ac

[2(a+1)]

2

=4(a

2

)(4)

4(a+1)

2

=4(a

2

)(4)

a

2

+1+2a=4a

2

3a

2

−2a−1=0

(a−1)(3a+1)=0

a=1,−

3

1

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Answer 2

The value of n , the equation $2x^{2} -nx-+n=0$ has coincident roots are 0 and 8.

Step-by-step explanation:

Given:

The equation $2x^{2} -nx-+n=0$ has coincident roots​.

To Find:

The value of n , the equation $2x^{2} -nx-+n=0$ has coincident roots .

Formula Used:

Quardatic equation $ax^{2} +bx+c =0$.

It is discriminant $D=b^{2} -4ac$

For  both roots are coincident roots​ the discriminant D =0

Solution:

As given-The equation $2x^{2} -nx-+n=0$ has coincident roots​.

Compared with quardatic equation $ax^{2} +bx+c =0$.

a=2,   b= - n   and c =n

For  both roots are coincident roots​ the discriminant D =0

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

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

$0=(-n)^{2} -4\times2 \times n$

$0=n^{2} -8 n$

$0=n(n-8)$

$(n-8)=0$

$n=8$

$n=0$

Thus,the value of n , the equation $2x^{2} -nx-+n=0$ has coincident roots are 0 and 8.