Question

If both (x-2) and ( x-1/2) are the factors of px^2+5x+r show that p=r

Answer 1

Heya ✋

Let p(x) = px^2 + 5x + r

g(x) = (x - 2)

f(x) = (x - 1/2)

p(2) = p × 2^2 + 5 × 2 + r

= p × 4 + 10 + r

= 4p + 10 + r ......(i)

p(1/2) = p × (1/2)^2 + 5 × 1/2 + r

= p × 1/4 + 5/2 + r

= p/4 + 5/2 + r .......(ii)

From equation (i) and (ii)

4p + 10 + r = p/4 + 5/2 + r

=> 4p - p/4 + r - r = 5/2 - 10

=> 16p - p/4 = 5 - 20/2

=> 15p/4 = -15/2

=> p = -15/2 × 4/15

=> p = -2

Hence ,

4p + r = -10 [equation (i)]

=> 4 × (-2) + r = -10

=> -8 + r = -10

=> r = -10 + 8

=> r = -2

Therefore ,

p = r

proved







Thanks :)))))

Answer 2

Hey friend!!!✋ here is your answer

______________________________

◆To show - p=q

✔️If (x-2) & (x-1/2) is factor then

x-2 =0. & x-1/2=0

✅x=2. & x=1/2

✔️Put value in given equation

✔️x=2

✅p(2)^2+5(2)+r

✅4p+10+r------(1)

✔️Now x=1/2

✔️p(1/2)^2+5(1/2)+r

✅P/4+5/2+r-----(2)


✔️ By equation 1 & 2

✅4p+10+r=P/4+5/2+r

✔️4p+10+r= P+(5×2)+4r/4 {By taking LCM)

✅4p+10+r=p+10+4r/4
✅Cross multiply

✔️4(4p+10+r)=P+10+4r

✅16p+40+4r=P+10+4r

✔️16p-p=10-40

✅15p=-30

✔️P=-30/15

✅P=-2

◆Put Value of P in equation (1)

✔️4(-2)+10+r=0

✅-8+10+r=0

✔️2+r=0

✅r=-2

So p=r=-2

__Hence proved_______________

⭐️Hope it helps you⭐️