Question

if both (x-2) and (x-1/2) are the factors of px²+5x+r,show that p=r

Answer 1

Step-by-step explanation:

Given, f(x) = px^2+5x+r and factors are x-2, x-1/2

Substitute x = 2 in place of equation, we get

= p*2^2+5*2+r=0

= 4p + 10 + r = 0 ------------ (i)

Substitute x = 1/2 in place of equation.

p/4 + 5/2 + r = 0

p + 10 + 4r = 0 ------------------- (ii)

On solving (i),(ii) we get

4p+r=-10 and p+4r+10=0

4p+r=p+4r

3p=3r

• p = r.Given, f(x) = px^2+5x+r and factors are x-2, x-1/2
• Substitute x = 2 in place of equation, we get
• = p*2^2+5*2+r=0
• = 4p + 10 + r = 0 ------------ (i)
• Substitute x = 1/2 in place of equation.
• p/4 + 5/2 + r = 0
• p + 10 + 4r = 0 ------------------- (ii)
• On solving (i),(ii) we get
• 4p+r=-10 and p+4r+10=0
• 4p+r=p+4r
• 3p=3r
• p = r.

Answer 2

GIVEN:-

• if both (x-2) and (x-1/2) are the factors of px²+5x+r,

TO PROVE:-

• p = r.

Now,

$\implies\rm{ x - 2 = 0}$

$\implies\rm{ f(x) = 2}$

Putting the value of "x" in f(x).

$\implies\rm{ p(x) = px^2 + 5x +r}$

$\implies\rm{ p(2) = p(2)^2 + 5(2) + r }$

$\implies\rm{ 4p + 10 + r = 0}$

$\implies\rm{ 4p + r = -10}$..........eq1

Now,

$\implies\rm{ f(x) = x - \dfrac{1}{2} = 0}$

$\implies\rm{ x = \dfrac{1}{2}}$

Putting the Value of x in p(x).

$\implies\rm{ p(x) = p(\dfrac{1}{2})^2 + 5(\dfrac{1}{2}) + r }$

$\implies\rm{ \dfrac{P}{4} + \dfrac{5}{2} + r = 0}$

$\implies\rm{\dfrac{4p+ 10 + 4r}{4} = 0}$

$\implies\rm{ p + 10 + 4r =0}$

$\implies\rm{ p + 4r = -10}$.........eq2

Subtracting the eq.1 and eq 2.

$\implies\rm{ 4p + r -(p +4r) = -10-(-10)}$

$\implies\rm{ 4p + r - p -4r = 0}$

$\implies\rm{ 3p - 3r = 0}$

$\implies\rm{ 3p = 3r}$

$\implies\rm{ p = r}$.

Hence, Proved