Question

Class 9 Chapter 2 ex2.2 question 3​

Answer 1

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3. Verify whether the following are zeroes of the polynomial, indicated against them.

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(i) p(x)=3x+1, x=−1/3

Solution:

For, x = -1/3, p(x) = 3x+1

∴p(−1/3) = 3(-1/3)+1 = −1+1 = 0

∴ -1/3 is a zero of p(x).

(ii) p(x)=5x–π, x = 4/5

Solution:

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- = 4-

∴ 4/5 is not a zero of p(x).

(iii) p(x)=x2−1, x=1, −1

Solution:

For, x = 1, −1;

p(x) = x2−1

∴p(1)=12−1=1−1 = 0

p(−1)=(-1)2−1 = 1−1 = 0

∴1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2

Solution:

For, x = −1,2;

p(x) = (x+1)(x–2)

∴p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴−1,2 are zeros of p(x).

(v) p(x) = x2, x = 0

Solution:

For, x = 0 p(x) = x2

p(0) = 02 = 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx+m, x = −m/l

Solution:

For, x = -m/l ; p(x) = lx+m

∴ p(-m/l)= l(-m/l)+m = −m+m = 0

∴-m/l is a zero of p(x).

(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3

Solution:

For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1

∴p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0

∴p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0

∴-1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x).

(viii) p(x) =2x+1, x = 1/2

Solution:

For, x = 1/2 p(x) = 2x+1

∴ p(1/2)=2(1/2)+1 = 1+1 = 2≠0

∴1/2 is not a zero of p(x).

Answer 2

Answer:

(i) p(x)=3x+1, x=−1/3

Solution:

For, x = -1/3, p(x) = 3x+1

∴p(−1/3) = 3(-1/3)+1 = −1+1 = 0

∴ -1/3 is a zero of p(x).

(ii) p(x)=5x–π, x = 4/5

Solution:

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- = 4-

∴ 4/5 is not a zero of p(x).

(iii) p(x)=x2−1, x=1, −1

Solution:

For, x = 1, −1;

p(x) = x2−1

∴p(1)=12−1=1−1 = 0

p(−1)=(-1)2−1 = 1−1 = 0

∴1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2

Solution:

For, x = −1,2;

p(x) = (x+1)(x–2)

∴p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴−1,2 are zeros of p(x).

(v) p(x) = x2, x = 0

Solution:

For, x = 0 p(x) = x2

p(0) = 02 = 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx+m, x = −m/l

Solution:

For, x = -m/l ; p(x) = lx+m

∴ p(-m/l)= l(-m/l)+m = −m+m = 0

∴-m/l is a zero of p(x).

(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3

Solution:

For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1

∴p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0

∴p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0

∴-1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x).

(viii) p(x) =2x+1, x = 1/2

Solution:

For, x = 1/2 p(x) = 2x+1

∴ p(1/2)=2(1/2)+1 = 1+1 = 2≠0

∴1/2 is not a zero of p(x).