Class 9 Exam > Class 9 Questions > a)2p = rb)p = 2rc)p = rd)none of theseCorrect...
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- a)2p = r
- b)p = 2r
- c)p = r
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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a)2p = rb)p = 2rc)p = rd)none of theseCorrect answer is option 'C'. Ca...
Let f(x) = px2 + 5 x + r
If (x - 2) is a factor of f (x), then by factor theorem
f(2) = 0 | x - 2 = 0 ⇒ x = 2
⇒ p(2)2 + 5(2) + r = 0
⇒ 4p + r + 10 = 0 ...(1)
If (x - 2) is a factor of f (x), then by factor theorem
f(2) = 0 | x - 2 = 0 ⇒ x = 2
⇒ p(2)2 + 5(2) + r = 0
⇒ 4p + r + 10 = 0 ...(1)
If 
is a factor of f (x), then by factor theorem,
is a factor of f (x), then by factor theorem,
Subtracting (2) from (1), we get
3p - 3r = 0
⇒ p = r
3p - 3r = 0
⇒ p = r
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a)2p = rb)p = 2rc)p = rd)none of theseCorrect answer is option 'C'. Ca...
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a)2p = rb)p = 2rc)p = rd)none of theseCorrect answer is option 'C'. Ca...
P (x) = px^2 + 5x +r g (x) = x-2, x- 1/2 g (x) =0 g (x) = 0 x-2 =0 _ x= 2 x-1/2=0 _ x=1/2 now ,p (2): p (2)^2 +5×2+ r=0 -》4p + 10 +r = 0 4p+r = -10 (1) equation.
Now, p (1/2)= p (1/2)^2 + 5×1/2 + r =0 p/4 + 5/2 +r = 0 -》p/4 +r = -5/2 [ p+4r = -10 (2) EQUATION ] × 4 "" 4p + 16 r = -40 subtract equation 1 and 2 we get : -15r = 30 _ r= -2 find the value of p by equation 1 : 4p +r =-10 4p -2 = - 10 -》 2 ( 2p - 1 ) = -10 -》 2p - 1 = -5 2p = -5 +1 -》 2p = -4 -》 p = -2 hence , p= -2 r = -2 》 p= r.
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a)2p = rb)p = 2rc)p = rd)none of theseCorrect answer is option 'C'. Can you explain this answer?
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