Question

SAMPLE PAPER (2020-21)
CLASS-x
SUBJECT-MATHEMATICS
Time: 3 hrs
PART-A (Section-1)
Q. 1. What is the H.C.F. of two consecutive even numbers?
दो क्रमागत सम संख्याओं का H.C.F. ज्ञात कीजिए।
Q. 2. If the sum of the zeroes of the polynomial p(x) = kx²+2x+3k is equal to their product, then
find two value of *k'.
यदि बहुपद p(x) =kx+2x+3k के शून्यांको का योग उनके गुणनफल के बराबर हो तो 'K' का
मान ज्ञात कीजिए।
incident​

Answer 1

Explanation:

एचसीएफ ऑफ टू कंसेक्युटिव इवन नंबर इज आल्सो इवन नंबर पर कॉल ऑलवेज टॉक टू कंसेक्युटिव नंबर्स अरे इवन नंबर

Answer 2

SOLUTION

TO DETERMINE

1. The H.C.F. of two consecutive even numbers

2. If the sum of the zeroes of the below polynomial

$\sf{p(x) = k {x}^{2} + 2x + 3k }$

is equal to their product, then the value of k

EVALUATION

1. Suppose two consecutive even numbers are 2a and 2a + 2

Now

2a = 2× a

2a + 2 = 2× ( a + 1 )

Hence the required HCF = 2

For example take two consecutive even numbers 8 , 10

Then their HCF = 2

The H.C.F. of two consecutive even numbers = 2

2. Here the given Quadratic polynomial is

$\sf{p(x) = k {x}^{2} + 2x + 3k }$

Comparing with the quadratic polynomial

$\sf{a {x}^{2} + bx + c } \: \: \: \: \: \: we \: get$

a = k , b = 2 , c = 3k

Sum of the zeroes

$\displaystyle \sf{ = - \frac{b}{a} }$

$\displaystyle \sf{ = - \frac{2}{k} }$

Product of the Zeroes

$\displaystyle \sf{ = \frac{c}{a} }$

$\displaystyle \sf{ = \frac{3k}{k} }$

$= 3$

Now it is given that sum of the zeroes of the polynomial is equal to their product

So by the given condition

$\displaystyle \sf{ - \frac{2}{k} = 3 }$

$\displaystyle \sf{ \implies k = - \frac{2}{3} }$

Hence the required value is

$\displaystyle \sf{ k = - \frac{2}{3} }$

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