Question

2x^2-(p+1)x+(p-1)=0.if α-β=αβ,then what is the value of p

Answer 1

Question

2x²-(p+1)X+(p-1)=0 if $\alpha-\beta$=-$\alpha \beta$

Given

2x²-(p+1)x+(p-1)=0

$\alpha-\beta$=-$\alpha \beta$

To Find

p=?

Solution

Let the zeroes of the polynomial be $\alpha \beta$

Then,

$\alpha+\beta$=$\frac{-b}{a}$=$\frac{p+1}{2}$

$\alpha \beta$=$\frac{c}{a}$=$\frac{p-1}{2}$

Now,we have to find

($\alpha-\beta$)²

=>($\alpha-\beta$)²=($\alpha+\beta$)²-4$\alpha \beta$

=>($\alpha-\beta$)²=$\frac{p+1}{2}^2$-4×$\frac{p-1}{2}$

=>($\alpha-\beta$)²=$\frac{p+1}{2}^2$-2p+2

=>($\alpha-\beta$)²=$\frac{p^2+1+2p}{4}$-2p-2

=>($\alpha-\beta$)²=$\frac{p^2+1+2p-8p+8}{4}$

=>($\alpha-\beta$)²=$\frac{p^2+9-6p}{4}$

=>($\alpha-\beta$)²=($\frac{p-3}{2}$)²

=>($\alpha-\beta$)=($\frac{p+3}{2}$)

Now from given

$\alpha-\beta$ =-$\alpha \beta$

putting the values

=>$\frac{p+3}{2}$=($\frac{p-1}{2}$)

=>p+3=-p+1

=>2p=-2

$\because$=-1

$\boxed{\boxed{AnsweR=-1}}$

Answer 2

Answer:

Given,

2x^2-(p+1)x+(p-1)=0

where,

a-b=ab

we have to find,

p=?

Step-by-step explanation:

let the zeros of the polynomial be ab

then,

a+b=-x/y=(p+1)/2

ab=z/y=(p-1)/2

now,

(a-b)^2=(a+b)^2-4ab

= (p+1/2)^2-2(p-1)

=(p^2+1+2p)/4-(2p-2)

=(p^2+1+2p)/4-2p+2

=(p^2+1+2p-8p+8)/4

=(p^2-6p+9)/4

=> (a-b)=√(p^2-6p+9)/4

we know that,

a-b=ab

then,

√(p^2-6p+9)/4=(p-1)/2

after squaring both the side then we get,

=>(p^2-6p+9)/4=[(p-1)/2]^2

=>(p^2-6p+9)/4=(p^2-2p+1)/4

=>p^2-6p+9=p^2-2p+1

=>-6p+2p=1-9

=>-4p=-8

=>p=8/4=2

so,

the value of p is,

p=2