Your cart is empty.
Practice each question, then open the answer or worked solution.
If one of the zeroes of the quadratic polynomial (k–1) x2 + k x + 1 is –3, then the value of k is
(A) 4/3 (B) -4/3
(A) 4/3
Explanation:
According to the question,
-3 is one of the zeros of quadratic polynomial (k-1)x2+kx+1
Substituting -3 in the given polynomial,
(k-1)(-3)²+k(-3)+1=0
(k-1)9+k(-3)+1 = 0
9k-9-3k+1=0
6k-8=0
k=8/6
Therefore, k=4/3
Hence, option (A) is the correct answer.
A quadratic polynomial, whose zeroes are –3 and 4, is
(A) x2 – x + 12 (B) x2 + x + 12
(C) (x2/2)-(x/2)-6 (D) 2x2 + 2x –24
(C) (x2/2)-(x/2)-6
Explanation:
Sum of zeroes, α+ β= -3+4 =1
Product of Zeroes, αβ = -3× 4 = -12
Therefore, the quadratic polynomial becomes,
x²- (sum of zeroes)x+(product of zeroes)
= x²- (α+ β)x+(αβ)
= x² – (1)x + (-12)
= x² – x -12
divide by 2, we get
= x²/2 – x/2 -12/2
= x²/2 – x/2 -6
Hence, option (C) is the correct answer.
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
(D) a = 2, b = – 6
Explanation:
According to the question,
x² + (a+1)x + b
Given that, the zeroes of the polynomial = 2 and -3,
When x = 2
2² + (a+1)(2) + b = 0
4 + 2a+2 + b = 0
6 + 2a+b = 0
2a+b = -6 —– (1)
When x = -3,
(-3)² + (a+1)(-3) + b = 0
9 – 3a-3 + b = 0
6 – 3a+b = 0
-3a+b = -6 —– (2)
Subtracting equation (2) from (1)
2a+b – (-3a+b) = -6-(-6)
2a+b+3a-b = -6+6
5a = 0
a = 0
Substituting the value of ‘a’ in equation (1), we get,
2a + b = -6
2(0) +b = -6
b = -6
Hence, option (D) is the correct answer.
The number of polynomials having zeroes as –2 and 5 is
(A) 1 (B) 2
(C) 3 (D) more than 3
(D) more than 3
Explanation:
According to the question,
The zeroes of the polynomials = -2 and 5
We know that the polynomial is of the form,
p(x) = ax2 + bx + c.
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2 i.e.
Sum of the zeroes = – b/a
– 2 + 5 = – b/a
3 = – b/a
b = – 3 and a = 1
Product of the zeroes = constant term ÷ coefficient of x2 i.e.
Product of zeroes = c/a
(- 2)5 = c/a
– 10 = c
Substituting the values of a, b and c in the polynomial p(x) = ax2 + bx + c.
We get, x2 – 3x – 10
Therefore, we can conclude that x can take any value.
Hence, option (D) is the correct answer.
Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is
(A) (–c/a) (B) c/a
(C) 0 (D) (–b/a)
(B) (c/a)
Explanation:
According to the question,
We have the polynomial,
ax3 + bx2 + cx + d
We know that,
Sum of product of roots of a cubic equation is given by c/a
It is given that one root = 0
Now, let the other roots be α, β
So, we get,
αβ + β(0) + (0)α = c/a
αβ = c/a
Hence the product of other two roots is c/a
Hence, option (B) is the correct answer