If both x-2 and x-1-2 are factors of px2-5x-r - then -

Px^2+5x+r=0—(1) (x 2) ( x 1/2 )=0 [since it is factors] x^2 x/2 2x+1=0 x^2 ( x+4x/2 )+1=0 2x^2 5x+2=0 nbsp; nbsp; nbsp;—(2) Comparing equation(1) and eq ...

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Standard IX

Mathematics

Factor of Polynomials

If both x-2 a...

Question

If both $x - 2$ and $x - \frac{1}{2}$ are factors of $p x^{2} + 5 x + r$ , then .

p = r

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p + r = 0

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p + 2 r = 0

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Solution

The correct option is A $p = r$
$p x^{2} + 5 x + r = 0 - - - (1)$
$(x - 2) (x - \frac{1}{2}) = 0$
[since it is factors]
$x^{2} - \frac{x}{2} - 2 x + 1 = 0$
$x^{2} - (\frac{x + 4 x}{2}) + 1 = 0$
$2 x^{2} - 5 x + 2 = 0 - - - (2)$
Comparing equation(1) and equation(2), we get
$p = r$

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Similar questions

Q. Question 3
if both x – 2 and

x - \frac{1}{2}

are factors of

p x^{2} + 5 x + r

, then show that p = r.

Q. If both

(x - 2)

and

(x - \frac{1}{2})

are factors of

p x^{2} + 5 x + r

, show that

p = r

Q. If both

x - 2

and

x - \frac{1}{2}

are factors of

p x^{2} + 5 x + r

. Which of the conditions hold true?

Q. If both

x - 2

and

x - \frac{1}{2}

are factors of

p x^{2} + 5 x + r

, then

Q. If both x − 2 and

x - \frac{1}{2}

are factors of px² + 5x + r, then

(a) p = r

(b) p + r = 0

(c) 2p + r = 0

(d) p + 2r = 0

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If both x−2 and x−12 are factors of px2+5x+r , then .

The correct option is A p=rpx2+5x+r=0−−−(1) (x−2)(x−12)=0 [since it is factors] x2−x2−2x+1=0 x2−(x+4x2)+1=0 2x2−5x+2=0−−−(2) Comparing equation(1) and equation(2), we get p=r

If both $x - 2$ and $x - \frac{1}{2}$ are factors of $p x^{2} + 5 x + r$ , then .

The correct option is A $p = r$
$p x^{2} + 5 x + r = 0 - - - (1)$
$(x - 2) (x - \frac{1}{2}) = 0$
[since it is factors]
$x^{2} - \frac{x}{2} - 2 x + 1 = 0$
$x^{2} - (\frac{x + 4 x}{2}) + 1 = 0$
$2 x^{2} - 5 x + 2 = 0 - - - (2)$
Comparing equation(1) and equation(2), we get
$p = r$