If the hanging wire touches the bridge at only one point, then the
polynomial representing hanging wire will have to non
(a) no zeroes (b) two distinct zeroes (c) two equal zeroes
(d) three zeroes
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Answers
Given : hanging wire touches the bridge at only one point,
To Find : the polynomial representing hanging wire will have
(a) no zeroes
(b) two distinct zeroes
(c) two equal zeroes
(d) three zeroes
Solution:
if hanging wire does not touch or intersect the bridge at any point
then No zeroes
y = x² + 3 or - 1 - x²
if hanging wire intersect the bridge at one point
then One Zero
y = x + 2
hanging wire intersect the bridge at 2 points
2 distinct zeroes
y = (x - 2)(x - 3)
hanging wire touches the bridge at only one point,
Hence two Equal zero
Example : y = x² touch the x axis at ( 0 , 0) has two zeroes and both are 0 , 0 .
(d) three zeroes not possible as it touches at point so zeroes can be even number only
like 2 , 4 , 6
example y = x⁴ has 4 Equal zeroes.
Hence If the hanging wire touches the bridge at only one point, then the
polynomial representing hanging wire will have TWO EQUAL ZREOES
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