89973
If the zeroes of the quadratic polynomial x\(^{1}\) + (a + 1)x + b are 2 and -3, then
1 a = -7,b = -1
2 a = 5, b = -1
3 a = 2, b = -6
4 a = 0, b = -6
Explanation:
a = 0, b = -6 The given quadratic equation is x\(^{1}\) + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, \(\alpha=2\) and \(\beta=-3\) Now, Sum of zeroes \(=-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow2+(-3)=-\frac{(\text{a}+1)}{1}\) \(\Rightarrow-1=-\text{a}-1\) \(\Rightarrow\text{a}=0\) Product of zeroes \(=\frac{\text{Constant of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow 2\times(-3)=\frac{\text{b}}{1}\) \(\Rightarrow\text{b}=-6\) So, a = 0 and b = -6 Hence, the correct answer is option (d)
POLYNOMIALS
89974
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:
1 x\(^{1}\) - 9
2 x\(^{1}\) + 9
3 x\(^{1}\) + 3
4 x\(^{1}\) - 3
Explanation:
x\(^{1}\) - 9 Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomials such that \(0=\alpha+\beta\) If one of zero is 3 then \(\alpha+\beta=0\) \(3+\beta=0\) \(\beta=0-3\) \(\beta=-3\) Substituting \(\beta=-3\) in \(\alpha+\beta=0\) we get \(\alpha-3=0\) \(\alpha=3\) Let S and P denote the sum and product of the zeros of the polynomial respectively then \(\text{S}=\alpha+\beta\) \(\text{S}=0\) \(\text{p}=\alpha\beta\) \(\text{p}=3\times-3\) \(\text{p}=-9\) Hence, the required polynomials is \(=(\text{x}^2-\text{Sx}+\text{p})\) \(=(\text{x}^2-0\text{x}-9)\) \(=\text{x}^2-9\) Hence, the correct choice is (a)
POLYNOMIALS
89975
If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then the value of k is:
1 6
2 -6
3 5
4 -5
Explanation:
-6 If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3)\(^{1}\) - 4(3)\(^{1}\) - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6
POLYNOMIALS
89976
If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, then \(\alpha +\beta = \)
1 \(\frac{\text{c}}{\text{a}}\)
2 \(\frac{\text{b}}{\text{a}}\)
3 \(\frac{-\text{b}}{\text{a}}\)
4 \(\frac{-\text{c}}{\text{a}}\)
Explanation:
\(\frac{-\text{b}}{\text{a}}\) If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, \(\because\) Sum of the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c \(=\frac{\text{-(Coe f ficient of x)}}{\text{Coe f ficient of x}^{2}} \text{ then } \alpha+\beta = \frac{-b}{a}\)
89973
If the zeroes of the quadratic polynomial x\(^{1}\) + (a + 1)x + b are 2 and -3, then
1 a = -7,b = -1
2 a = 5, b = -1
3 a = 2, b = -6
4 a = 0, b = -6
Explanation:
a = 0, b = -6 The given quadratic equation is x\(^{1}\) + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, \(\alpha=2\) and \(\beta=-3\) Now, Sum of zeroes \(=-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow2+(-3)=-\frac{(\text{a}+1)}{1}\) \(\Rightarrow-1=-\text{a}-1\) \(\Rightarrow\text{a}=0\) Product of zeroes \(=\frac{\text{Constant of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow 2\times(-3)=\frac{\text{b}}{1}\) \(\Rightarrow\text{b}=-6\) So, a = 0 and b = -6 Hence, the correct answer is option (d)
POLYNOMIALS
89974
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:
1 x\(^{1}\) - 9
2 x\(^{1}\) + 9
3 x\(^{1}\) + 3
4 x\(^{1}\) - 3
Explanation:
x\(^{1}\) - 9 Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomials such that \(0=\alpha+\beta\) If one of zero is 3 then \(\alpha+\beta=0\) \(3+\beta=0\) \(\beta=0-3\) \(\beta=-3\) Substituting \(\beta=-3\) in \(\alpha+\beta=0\) we get \(\alpha-3=0\) \(\alpha=3\) Let S and P denote the sum and product of the zeros of the polynomial respectively then \(\text{S}=\alpha+\beta\) \(\text{S}=0\) \(\text{p}=\alpha\beta\) \(\text{p}=3\times-3\) \(\text{p}=-9\) Hence, the required polynomials is \(=(\text{x}^2-\text{Sx}+\text{p})\) \(=(\text{x}^2-0\text{x}-9)\) \(=\text{x}^2-9\) Hence, the correct choice is (a)
POLYNOMIALS
89975
If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then the value of k is:
1 6
2 -6
3 5
4 -5
Explanation:
-6 If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3)\(^{1}\) - 4(3)\(^{1}\) - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6
POLYNOMIALS
89976
If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, then \(\alpha +\beta = \)
1 \(\frac{\text{c}}{\text{a}}\)
2 \(\frac{\text{b}}{\text{a}}\)
3 \(\frac{-\text{b}}{\text{a}}\)
4 \(\frac{-\text{c}}{\text{a}}\)
Explanation:
\(\frac{-\text{b}}{\text{a}}\) If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, \(\because\) Sum of the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c \(=\frac{\text{-(Coe f ficient of x)}}{\text{Coe f ficient of x}^{2}} \text{ then } \alpha+\beta = \frac{-b}{a}\)
89973
If the zeroes of the quadratic polynomial x\(^{1}\) + (a + 1)x + b are 2 and -3, then
1 a = -7,b = -1
2 a = 5, b = -1
3 a = 2, b = -6
4 a = 0, b = -6
Explanation:
a = 0, b = -6 The given quadratic equation is x\(^{1}\) + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, \(\alpha=2\) and \(\beta=-3\) Now, Sum of zeroes \(=-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow2+(-3)=-\frac{(\text{a}+1)}{1}\) \(\Rightarrow-1=-\text{a}-1\) \(\Rightarrow\text{a}=0\) Product of zeroes \(=\frac{\text{Constant of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow 2\times(-3)=\frac{\text{b}}{1}\) \(\Rightarrow\text{b}=-6\) So, a = 0 and b = -6 Hence, the correct answer is option (d)
POLYNOMIALS
89974
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:
1 x\(^{1}\) - 9
2 x\(^{1}\) + 9
3 x\(^{1}\) + 3
4 x\(^{1}\) - 3
Explanation:
x\(^{1}\) - 9 Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomials such that \(0=\alpha+\beta\) If one of zero is 3 then \(\alpha+\beta=0\) \(3+\beta=0\) \(\beta=0-3\) \(\beta=-3\) Substituting \(\beta=-3\) in \(\alpha+\beta=0\) we get \(\alpha-3=0\) \(\alpha=3\) Let S and P denote the sum and product of the zeros of the polynomial respectively then \(\text{S}=\alpha+\beta\) \(\text{S}=0\) \(\text{p}=\alpha\beta\) \(\text{p}=3\times-3\) \(\text{p}=-9\) Hence, the required polynomials is \(=(\text{x}^2-\text{Sx}+\text{p})\) \(=(\text{x}^2-0\text{x}-9)\) \(=\text{x}^2-9\) Hence, the correct choice is (a)
POLYNOMIALS
89975
If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then the value of k is:
1 6
2 -6
3 5
4 -5
Explanation:
-6 If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3)\(^{1}\) - 4(3)\(^{1}\) - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6
POLYNOMIALS
89976
If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, then \(\alpha +\beta = \)
1 \(\frac{\text{c}}{\text{a}}\)
2 \(\frac{\text{b}}{\text{a}}\)
3 \(\frac{-\text{b}}{\text{a}}\)
4 \(\frac{-\text{c}}{\text{a}}\)
Explanation:
\(\frac{-\text{b}}{\text{a}}\) If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, \(\because\) Sum of the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c \(=\frac{\text{-(Coe f ficient of x)}}{\text{Coe f ficient of x}^{2}} \text{ then } \alpha+\beta = \frac{-b}{a}\)
89973
If the zeroes of the quadratic polynomial x\(^{1}\) + (a + 1)x + b are 2 and -3, then
1 a = -7,b = -1
2 a = 5, b = -1
3 a = 2, b = -6
4 a = 0, b = -6
Explanation:
a = 0, b = -6 The given quadratic equation is x\(^{1}\) + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, \(\alpha=2\) and \(\beta=-3\) Now, Sum of zeroes \(=-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow2+(-3)=-\frac{(\text{a}+1)}{1}\) \(\Rightarrow-1=-\text{a}-1\) \(\Rightarrow\text{a}=0\) Product of zeroes \(=\frac{\text{Constant of x}}{\text{Coefficient of x}^2}\) \(\Rightarrow 2\times(-3)=\frac{\text{b}}{1}\) \(\Rightarrow\text{b}=-6\) So, a = 0 and b = -6 Hence, the correct answer is option (d)
POLYNOMIALS
89974
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:
1 x\(^{1}\) - 9
2 x\(^{1}\) + 9
3 x\(^{1}\) + 3
4 x\(^{1}\) - 3
Explanation:
x\(^{1}\) - 9 Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomials such that \(0=\alpha+\beta\) If one of zero is 3 then \(\alpha+\beta=0\) \(3+\beta=0\) \(\beta=0-3\) \(\beta=-3\) Substituting \(\beta=-3\) in \(\alpha+\beta=0\) we get \(\alpha-3=0\) \(\alpha=3\) Let S and P denote the sum and product of the zeros of the polynomial respectively then \(\text{S}=\alpha+\beta\) \(\text{S}=0\) \(\text{p}=\alpha\beta\) \(\text{p}=3\times-3\) \(\text{p}=-9\) Hence, the required polynomials is \(=(\text{x}^2-\text{Sx}+\text{p})\) \(=(\text{x}^2-0\text{x}-9)\) \(=\text{x}^2-9\) Hence, the correct choice is (a)
POLYNOMIALS
89975
If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then the value of k is:
1 6
2 -6
3 5
4 -5
Explanation:
-6 If the polynomial 3x\(^{1}\) - 4x\(^{1}\) - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3)\(^{1}\) - 4(3)\(^{1}\) - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6
POLYNOMIALS
89976
If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, then \(\alpha +\beta = \)
1 \(\frac{\text{c}}{\text{a}}\)
2 \(\frac{\text{b}}{\text{a}}\)
3 \(\frac{-\text{b}}{\text{a}}\)
4 \(\frac{-\text{c}}{\text{a}}\)
Explanation:
\(\frac{-\text{b}}{\text{a}}\) If \(\alpha\) and \(\beta\) are the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c, \(\because\) Sum of the zeroes of a quadratic polynomial ax\(^{1}\) + bx + c \(=\frac{\text{-(Coe f ficient of x)}}{\text{Coe f ficient of x}^{2}} \text{ then } \alpha+\beta = \frac{-b}{a}\)
