Question

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

Answer 1

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.

Hope this helps!

Answer 2

$\red {★★★★☆☆☆✭✭✮✭✭☆☆☆★★★★}$

$\bf f(x) = p {x}^{2} + 5x + r$

$\bf f(2) = p(2 {)}^{2} + {5}^{2} + r$

$\bf p(4) + 10 + r = 0 \\ \bf 4p + 10 + r = 0 \\ \bf 4p + r = 0 - 10 \\ \bf 4p + r = - 10⇒(1)$

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$\bf f(x) = p {x}^{2} + 5x + r \\ \bf f( \dfrac{1}{2}) = p( \dfrac{1}{2}) + r = 0 \\ \bf p( \dfrac{1}{4} ) + \dfrac{5}{2} + r = 0$

$\bf \dfrac{p + 10 + 4r}{4} = 0 \\ \bf p + 10 + 4r = 0(4) \\ \bf p +4r = 0 - 10 \\ \bf p + 4r = - 10⇒(2)$

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$\bf From \: (1) \: and \: (2)$

$\bf 4p + r = - 10(1) \\ \bf p + 4r = - 10(2)$

$\bf 4p + r = p + 4r \\ \bf 4p - p = 4r - r \\ \bf 3p = 3r \\ \bf p = \dfrac{3r}{3} \\ \therefore p = r$

$\red{★★★★☆☆☆✭✭✮✮✮☆☆☆★★★★}$