90018
A quadratic polynomial whose product and sum of zeroes are \(\frac{1}{3}\) and \(\sqrt{2}\) respectively is:
1 \(\text{3x}^{2}+\text{x }-3\sqrt{2}\text{x}\)
2 \(\text{3x}^{2}-\text{x+3 }\sqrt{2}\text{x}\)
3 \(\text{3x}^{2}+3 \sqrt{ 2}\text{x}+{1}\)
4 \(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\)
Explanation:
\(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\) Given: \(\alpha+\beta=\frac{\sqrt{2}}{1} =\frac{ (-\sqrt{2})}{1} = \frac{(-3\sqrt{2})}{3}\) And \( \alpha\beta = \frac{\text{c}}{\text{a}} = \frac{1}{3}\) on comparing, we get \(\text{a = 3, b = }-3\sqrt{2}, \text{c} = {1}\) Putting these values in the general form of a quadratic polynomial ax\(^{1}\) + bx + c, we have \({3}\text{x}^{2} - {3}\sqrt{2}+{1}\)
POLYNOMIALS
90020
If the zeroes of a quadratic polynomial ax\(^{1}\) + bc + c, c ≠ 0 are equal, then:
1 C and a have opposite signs.
2 C and b have opposite signs.
3 C and a have the same sign.
4 C and b have the same sign.
Explanation:
C and a have the same sign. Let the given quadratic polynomial be f(x) = ax\(^{1}\) + bx + c Suppose \(\alpha\) and \(\beta\) be the zeroes of the given polynomial. Since \(\alpha\) and \(\beta\) are equal so they will have the same sign i.e., either both are positive or both are negative. So, \(\alpha\beta>0\) But \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\therefore\ \frac{\text{c}}{\text{a}}>0,\) which is possible only when both have same sign Hence, the correct answer is option (c)
POLYNOMIALS
90021
If \(\alpha,\beta\) are the zeros of the polynomial p(x) = 4x\(^{1}\) + 3x + 7, then \(\frac{1}{\alpha}+\frac{1}{\beta}\) is equal to:
1 \(\frac{7}{3}\)
2 \(\frac{-7}{3}\)
3 \(\frac{3}{7}\)
4 \(\frac{-3}{7}\)
Explanation:
\(\frac{-3}{7}\) Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomial p(x) = 4x\(^{1}\) + 3x + 7 \(\alpha+\beta=-\frac{\text{coefficient of x}}{\text{coefficient of x}^2}\) \(=\frac{-3}{4}\) \(\alpha\beta=\frac{\text{constant term}}{\text{coefficient of x}^2}\) \(=\frac{7}{4}\) We have \(=\frac{1}{\alpha}+\frac{1}{\beta}\) \(=\frac{\beta+\alpha}{\alpha\beta}\) \(=\frac{\frac{-3}{4}}{\frac{7}{4}}\) \(=\frac{-3}{4}\times\frac{4}{7}\) \(=\frac{-3}{7}\) The value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is \(\frac{-3}{7}\) Hence, the correct choice is (d).
POLYNOMIALS
90022
if \(\alpha\) and \(\beta\) are the zeroes of the polynomial 3x\(^{1}\) + 11x - 4, then the value of \(\frac{1}\alpha+\frac{1}\beta\) is:
1 \(\frac{11}{4}\)
2 \(\frac{12}{4}\)
3 \(\frac{13}{4}\)
4 \(\frac{15}{4}\)
Explanation:
\(\frac{11}{4}\) Here a = 3, b = 11, c = -4, \(\text{Since }\frac{1}\alpha+\frac{1}\beta = \frac{\alpha+\beta}{\alpha\beta}\) \(\alpha+\beta = \frac{-11}{3}, \alpha\beta= \frac{-4}{3}\) \(\text{So } \frac{-11}{3}\frac{-4}{3} = \frac{11}{4}\)
90018
A quadratic polynomial whose product and sum of zeroes are \(\frac{1}{3}\) and \(\sqrt{2}\) respectively is:
1 \(\text{3x}^{2}+\text{x }-3\sqrt{2}\text{x}\)
2 \(\text{3x}^{2}-\text{x+3 }\sqrt{2}\text{x}\)
3 \(\text{3x}^{2}+3 \sqrt{ 2}\text{x}+{1}\)
4 \(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\)
Explanation:
\(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\) Given: \(\alpha+\beta=\frac{\sqrt{2}}{1} =\frac{ (-\sqrt{2})}{1} = \frac{(-3\sqrt{2})}{3}\) And \( \alpha\beta = \frac{\text{c}}{\text{a}} = \frac{1}{3}\) on comparing, we get \(\text{a = 3, b = }-3\sqrt{2}, \text{c} = {1}\) Putting these values in the general form of a quadratic polynomial ax\(^{1}\) + bx + c, we have \({3}\text{x}^{2} - {3}\sqrt{2}+{1}\)
POLYNOMIALS
90020
If the zeroes of a quadratic polynomial ax\(^{1}\) + bc + c, c ≠ 0 are equal, then:
1 C and a have opposite signs.
2 C and b have opposite signs.
3 C and a have the same sign.
4 C and b have the same sign.
Explanation:
C and a have the same sign. Let the given quadratic polynomial be f(x) = ax\(^{1}\) + bx + c Suppose \(\alpha\) and \(\beta\) be the zeroes of the given polynomial. Since \(\alpha\) and \(\beta\) are equal so they will have the same sign i.e., either both are positive or both are negative. So, \(\alpha\beta>0\) But \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\therefore\ \frac{\text{c}}{\text{a}}>0,\) which is possible only when both have same sign Hence, the correct answer is option (c)
POLYNOMIALS
90021
If \(\alpha,\beta\) are the zeros of the polynomial p(x) = 4x\(^{1}\) + 3x + 7, then \(\frac{1}{\alpha}+\frac{1}{\beta}\) is equal to:
1 \(\frac{7}{3}\)
2 \(\frac{-7}{3}\)
3 \(\frac{3}{7}\)
4 \(\frac{-3}{7}\)
Explanation:
\(\frac{-3}{7}\) Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomial p(x) = 4x\(^{1}\) + 3x + 7 \(\alpha+\beta=-\frac{\text{coefficient of x}}{\text{coefficient of x}^2}\) \(=\frac{-3}{4}\) \(\alpha\beta=\frac{\text{constant term}}{\text{coefficient of x}^2}\) \(=\frac{7}{4}\) We have \(=\frac{1}{\alpha}+\frac{1}{\beta}\) \(=\frac{\beta+\alpha}{\alpha\beta}\) \(=\frac{\frac{-3}{4}}{\frac{7}{4}}\) \(=\frac{-3}{4}\times\frac{4}{7}\) \(=\frac{-3}{7}\) The value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is \(\frac{-3}{7}\) Hence, the correct choice is (d).
POLYNOMIALS
90022
if \(\alpha\) and \(\beta\) are the zeroes of the polynomial 3x\(^{1}\) + 11x - 4, then the value of \(\frac{1}\alpha+\frac{1}\beta\) is:
1 \(\frac{11}{4}\)
2 \(\frac{12}{4}\)
3 \(\frac{13}{4}\)
4 \(\frac{15}{4}\)
Explanation:
\(\frac{11}{4}\) Here a = 3, b = 11, c = -4, \(\text{Since }\frac{1}\alpha+\frac{1}\beta = \frac{\alpha+\beta}{\alpha\beta}\) \(\alpha+\beta = \frac{-11}{3}, \alpha\beta= \frac{-4}{3}\) \(\text{So } \frac{-11}{3}\frac{-4}{3} = \frac{11}{4}\)
90018
A quadratic polynomial whose product and sum of zeroes are \(\frac{1}{3}\) and \(\sqrt{2}\) respectively is:
1 \(\text{3x}^{2}+\text{x }-3\sqrt{2}\text{x}\)
2 \(\text{3x}^{2}-\text{x+3 }\sqrt{2}\text{x}\)
3 \(\text{3x}^{2}+3 \sqrt{ 2}\text{x}+{1}\)
4 \(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\)
Explanation:
\(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\) Given: \(\alpha+\beta=\frac{\sqrt{2}}{1} =\frac{ (-\sqrt{2})}{1} = \frac{(-3\sqrt{2})}{3}\) And \( \alpha\beta = \frac{\text{c}}{\text{a}} = \frac{1}{3}\) on comparing, we get \(\text{a = 3, b = }-3\sqrt{2}, \text{c} = {1}\) Putting these values in the general form of a quadratic polynomial ax\(^{1}\) + bx + c, we have \({3}\text{x}^{2} - {3}\sqrt{2}+{1}\)
POLYNOMIALS
90020
If the zeroes of a quadratic polynomial ax\(^{1}\) + bc + c, c ≠ 0 are equal, then:
1 C and a have opposite signs.
2 C and b have opposite signs.
3 C and a have the same sign.
4 C and b have the same sign.
Explanation:
C and a have the same sign. Let the given quadratic polynomial be f(x) = ax\(^{1}\) + bx + c Suppose \(\alpha\) and \(\beta\) be the zeroes of the given polynomial. Since \(\alpha\) and \(\beta\) are equal so they will have the same sign i.e., either both are positive or both are negative. So, \(\alpha\beta>0\) But \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\therefore\ \frac{\text{c}}{\text{a}}>0,\) which is possible only when both have same sign Hence, the correct answer is option (c)
POLYNOMIALS
90021
If \(\alpha,\beta\) are the zeros of the polynomial p(x) = 4x\(^{1}\) + 3x + 7, then \(\frac{1}{\alpha}+\frac{1}{\beta}\) is equal to:
1 \(\frac{7}{3}\)
2 \(\frac{-7}{3}\)
3 \(\frac{3}{7}\)
4 \(\frac{-3}{7}\)
Explanation:
\(\frac{-3}{7}\) Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomial p(x) = 4x\(^{1}\) + 3x + 7 \(\alpha+\beta=-\frac{\text{coefficient of x}}{\text{coefficient of x}^2}\) \(=\frac{-3}{4}\) \(\alpha\beta=\frac{\text{constant term}}{\text{coefficient of x}^2}\) \(=\frac{7}{4}\) We have \(=\frac{1}{\alpha}+\frac{1}{\beta}\) \(=\frac{\beta+\alpha}{\alpha\beta}\) \(=\frac{\frac{-3}{4}}{\frac{7}{4}}\) \(=\frac{-3}{4}\times\frac{4}{7}\) \(=\frac{-3}{7}\) The value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is \(\frac{-3}{7}\) Hence, the correct choice is (d).
POLYNOMIALS
90022
if \(\alpha\) and \(\beta\) are the zeroes of the polynomial 3x\(^{1}\) + 11x - 4, then the value of \(\frac{1}\alpha+\frac{1}\beta\) is:
1 \(\frac{11}{4}\)
2 \(\frac{12}{4}\)
3 \(\frac{13}{4}\)
4 \(\frac{15}{4}\)
Explanation:
\(\frac{11}{4}\) Here a = 3, b = 11, c = -4, \(\text{Since }\frac{1}\alpha+\frac{1}\beta = \frac{\alpha+\beta}{\alpha\beta}\) \(\alpha+\beta = \frac{-11}{3}, \alpha\beta= \frac{-4}{3}\) \(\text{So } \frac{-11}{3}\frac{-4}{3} = \frac{11}{4}\)
90018
A quadratic polynomial whose product and sum of zeroes are \(\frac{1}{3}\) and \(\sqrt{2}\) respectively is:
1 \(\text{3x}^{2}+\text{x }-3\sqrt{2}\text{x}\)
2 \(\text{3x}^{2}-\text{x+3 }\sqrt{2}\text{x}\)
3 \(\text{3x}^{2}+3 \sqrt{ 2}\text{x}+{1}\)
4 \(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\)
Explanation:
\(\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}\) Given: \(\alpha+\beta=\frac{\sqrt{2}}{1} =\frac{ (-\sqrt{2})}{1} = \frac{(-3\sqrt{2})}{3}\) And \( \alpha\beta = \frac{\text{c}}{\text{a}} = \frac{1}{3}\) on comparing, we get \(\text{a = 3, b = }-3\sqrt{2}, \text{c} = {1}\) Putting these values in the general form of a quadratic polynomial ax\(^{1}\) + bx + c, we have \({3}\text{x}^{2} - {3}\sqrt{2}+{1}\)
POLYNOMIALS
90020
If the zeroes of a quadratic polynomial ax\(^{1}\) + bc + c, c ≠ 0 are equal, then:
1 C and a have opposite signs.
2 C and b have opposite signs.
3 C and a have the same sign.
4 C and b have the same sign.
Explanation:
C and a have the same sign. Let the given quadratic polynomial be f(x) = ax\(^{1}\) + bx + c Suppose \(\alpha\) and \(\beta\) be the zeroes of the given polynomial. Since \(\alpha\) and \(\beta\) are equal so they will have the same sign i.e., either both are positive or both are negative. So, \(\alpha\beta>0\) But \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\therefore\ \frac{\text{c}}{\text{a}}>0,\) which is possible only when both have same sign Hence, the correct answer is option (c)
POLYNOMIALS
90021
If \(\alpha,\beta\) are the zeros of the polynomial p(x) = 4x\(^{1}\) + 3x + 7, then \(\frac{1}{\alpha}+\frac{1}{\beta}\) is equal to:
1 \(\frac{7}{3}\)
2 \(\frac{-7}{3}\)
3 \(\frac{3}{7}\)
4 \(\frac{-3}{7}\)
Explanation:
\(\frac{-3}{7}\) Since \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomial p(x) = 4x\(^{1}\) + 3x + 7 \(\alpha+\beta=-\frac{\text{coefficient of x}}{\text{coefficient of x}^2}\) \(=\frac{-3}{4}\) \(\alpha\beta=\frac{\text{constant term}}{\text{coefficient of x}^2}\) \(=\frac{7}{4}\) We have \(=\frac{1}{\alpha}+\frac{1}{\beta}\) \(=\frac{\beta+\alpha}{\alpha\beta}\) \(=\frac{\frac{-3}{4}}{\frac{7}{4}}\) \(=\frac{-3}{4}\times\frac{4}{7}\) \(=\frac{-3}{7}\) The value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is \(\frac{-3}{7}\) Hence, the correct choice is (d).
POLYNOMIALS
90022
if \(\alpha\) and \(\beta\) are the zeroes of the polynomial 3x\(^{1}\) + 11x - 4, then the value of \(\frac{1}\alpha+\frac{1}\beta\) is:
1 \(\frac{11}{4}\)
2 \(\frac{12}{4}\)
3 \(\frac{13}{4}\)
4 \(\frac{15}{4}\)
Explanation:
\(\frac{11}{4}\) Here a = 3, b = 11, c = -4, \(\text{Since }\frac{1}\alpha+\frac{1}\beta = \frac{\alpha+\beta}{\alpha\beta}\) \(\alpha+\beta = \frac{-11}{3}, \alpha\beta= \frac{-4}{3}\) \(\text{So } \frac{-11}{3}\frac{-4}{3} = \frac{11}{4}\)
