zeroes of following polynomial :
1) . 2x²+ 7x + 3
2) . 3x² - x - 4
Answers
Answer:
भढदमडदड
तड़डथमडधश्रडरढढदढ
Step-by-step explanation:
Q1. The graphs of y = p(x) are given in the figure given below. for some polynomials p (x). Find the number of zeroes of p (x), in each case
Sol. (i) The given graph is parallel to x-axis. It does not intersect the x-axis.
• It has no zeroes.
(ii) The given graph intersects the x-axis at one point only.
• It has one zero.
(iii) The given graph intersects the x-axis at three points.
• It has three zeroes.
(iv) The given graph intersects the x-axis at two points.
• It has two zeroes.
(v) The given graph intersects the x-axis at four points.
• It has four zeroes.
(vi) The given graph meets the x-axis at three points.
• It has three zeroes.
NCERT TEXTBOOK QUESTIONS SOLVED
Exercise– 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Sol. (i) x2 – 2x – 8
We have p(x) = x2 – 2x – 8
= x2 + 2x – 4x – 8 = x (x + 2) – 4 (x + 2)
= (x – 4) (x + 2)
For p(x) = 0, we have
(x – 4) (x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = – 2
∴ The zeroes of x2 2x – 8 are 4 and –2.
Now, sum of the zeroes
Thus, relationship between zeroes and the coefficients in x2 – 2x – 8 is verified.
(ii) 4s2 – 4s + 1
We have p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1 = 2S (2S – 1) –1 (2s – 1)
= (2s – 1) (2s – 1)
For p(s) = 0, we have,
Now,
Thus, the relationship between the zeroes and coefficients in the polynomial 4s2 – 4s + 1 is verified.
(iii) 6x2 – 3 – 7x
We have
p (x) = 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x (2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
For p (x) = 0, we have,
Thus, the relationship between the zeroes and the coefficients in the polynomial 6x2 – 3 – 7x is verified.
(iv) 4u2 + 8u
We have, f(u) = 4u2 + 8u = 4u (u + 2)
For f(u) = 0,
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = –2
∴ The zeroes of 4u2 + 8u and 0 and –2
Now, 4u2 + 8u can be written as 4u2 + 8u + 0.
Thus, the relationship between the zeroes and the coefficients in the polynomial 4u2 + 8u is verified.
(v) t2 – 15
We have,
For f(t) = 0, we have
Now, we can write t2 – 15 as t2 + 0t – 15.
Thus, the relationship between the zeroes and the coefficients in the polynomial t2 – 15 is verified.
(vi) 3x2 – x – 4
We have,
f(x) = 3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x (x + 1) –4 (x + 1)
= (x + 1) (3x – 4)
For f(x) = 0 ⇒ (x + 1) (3x – 4) = 0
Either (x + 1) = 0 ⇒ x = – 1
Thus, the relationship between the zeroes and the coefficients in 3x2 – x – 4 is verified.
Q2. Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively:
Sol.
Note:
A quadratic polynomial whose zeroes are α and β is given by
p(x) = {x2 – (α + β) x + αβ}
p(x) = {x2 – (sum of the zeroes) x – (product of the zeroes)}
(i) Since, sum of the zeroes,
Product of the zeroes, α β = –1
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
Since, have same zeroes, is the required quadratic polynomial.
(ii) Since, sum of the zeroes,
Product of zeroes,
∴The required quadratic polynomial is
Since, have same zeroes, is required quadratic polynomial.
(iii) Since, sum of zeroes, (α + β) 0
Product of zeroes, αβ = 5
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
= x2 – (0) x + 5
= x2 + 5
(iv) Since, sum of the zeroes, (α + β) = 1
Product of the zeroes = 1
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
= x2 – (1) x + 1
= x2 – x + 1
(v) Since, sum of the zeroes,
Product of the zeroes
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
Since, and (4x2 + x + 1) have same zeroes, the required quadratic polynomial is (4x2 + x + 1).
(vi) Since, sum of the zeroes, (α + β) = 4
Product of the zeroes, αβ = 1
∴ The required quadratic polynomial is
x2 – (α + β) x + αβ
= x2 – (4)x + 1
= x2 – 4x + 1
Exercise– 2.3