A polynomial isn't as complicated as it sounds, because it's just an algebraic
expression with several terms. Usually, polynomials have more than one term, and
each term can be a variable, a number or some combination of variables and
numbers. Some people use polynomials in their heads every day without realizing it,
while others do it more consciously.
(i) A real number is called a zero of p(x), if p() = 0 . What will be the zeros of
p(x) = x2 – 2x – 3 are
(a) (1, -3)
(b) 3, - 1
(c) -3, - 1
(d) 1, 3
(ii) If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6
(iii) The zeroes of the quadratic polynomial x2 + 99x + 127 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
(iv) Zeroes of a polynomial can be expressed graphically. Number of zeroes of
polynomial is equal to the number of points where the graph of polynomial
(a) intersects x-axis
(b) intersect y – axis
(c) intersects y – axis or x-axis
(d) None of the above
(v) If 1 is one of the zeroes of the polynomial x² + x + k, then the value of k is:
(a) 2
(b) -2
(c) 4
(d) -4
Answers
Question (1) :- A real number is called a zero of p(x), if p(x) = 0 . What will be the zeros of p(x) = x2 – 2x – 3 are
(a) (1, -3)
(b) 3, - 1
(c) -3, - 1
(d) 1, 3
Solution :-
checking all options we get,
(a) (1, - 3)
→ f(x) = x² - 2x - 3
→ f(1) = (1)² - 2*1 - 3
→ f(1) = 1 - 2 - 3
→ f(1) = 1 - 5
→ f(1) = (-4) ≠ 0 .
therefore, option (a) and (d) are incorrect .
(b) 3 , -1
→ f(x) = x² - 2x - 3
→ f(3) = (3)² - 2*3 - 3
→ f(3) = 9 - 6 - 3
→ f(3) = 9 - 9
→ f(3) = 0 .
therefore, we can conclude that, (b) 3 , -1 will be the zeros of p(x) .
Question (2) :- the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6
Solution :-
→ f(x) = x² + (a + 1)x + b
→ f(2) = (2)² + (a + 1)2 + b
→ f(2) = 4 + 2a + 2 + b
→ 2a + b = (-6) -------------- Eqn.(1)
and,
→ f(x) = x² + (a + 1)x + b
→ f(-3) = (-3)² + (a + 1)(-3) + b
→ f(-3) = 9 - 3a - 3 + b
→ 3a - b = 6 -------------- Eqn.(2)
adding Eqn.(1) and Eqn.(2) ,
→ (2a + b) + (3a - b) = (-6) + 6
→ 2a + 3a + b - b = 0
→ 5a = 0
→ a = 0 .
Putting value of a in Eqn.(1),
→ 2 * 0 + b = (-6)
→ b = (-6) .
therefore, (D) a = 0, b = - 6 is correct answer .
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