89960
If \(\alpha,\ \beta\) are the zeros of kx\(^{1}\) - 2x + 3k such that \(\alpha+\beta=\alpha\beta,\) then k = ?
1 \(\frac{1}{3}\)
2 \(\frac{-1}{3}\)
3 \(\frac{2}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
\(\frac{2}{3}\) Here, p(x) = x\(^{1}\) - 2x + 3k Comparing the given polynomial with ax\(^{1}\)+ bx + c, we get: a = 1, b = -2 and c = 3k It is given that \(\alpha\) and \(\beta\) are the roots of the polynomial. \(\therefore\alpha+\beta=-\frac{\text{b}}{\text{a}}\) \(\Rightarrow\alpha+\beta=-\Big(\frac{-2}{1}\Big)\) \(\Rightarrow\alpha+\beta=2\dots(\text{i})\) Also, \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\Rightarrow\alpha\beta=\frac{\text{5k}}1{}\) \(\Rightarrow\alpha\beta=\text{3k}\dots(\text{ii})\) Now, \(\alpha+\beta=\alpha\beta\) \(\Rightarrow2=\text{3k}\) [Using (i) and (ii)] \(\Rightarrow\text{k}=\frac{2}{3}\)
POLYNOMIALS
89961
If \(\alpha\) and \(\beta\) are zeros of x\(^{1}\) + 5x + 8, then the value of \((\alpha+\beta)\) is:
1 8
2 5
3 -5
4 -8
Explanation:
-5 \(\text{x}^{2}+{5}\text{x}+{8}\) \(\alpha+\beta=\frac{\text{Coefficient of x}}{\text{Coefficient of }{\text{x}}^{2}}\) \(=\frac{-5}{1} = {-5}\)
\(\text{x}+\frac{3}{\text{x}}\) An expression of the form p(x) = a\(_{\1}\) + a\(_{\1}\)x + a\(_{\1}\)x\(^{1}\) + ... + a\(_{\1}\)x\(^{1}\), where a\(_{\1}\) ? 0, is called a polynomial in x of degree n. Here, a\(_{\1}\), a\(_{\1}\), a\(_{\1}\), ..., a\(_{\1}\) are real numbers and each power of x is a non-negative integer. Hence, \(\text{x}+\frac{3}{\text{x}}\) is not a polynomial.
POLYNOMIALS
89963
If the sum of the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 is 6, then the value of k is:
1 2
2 4
3 -2
4 -4
Explanation:
4 Let \(\alpha,\beta\) be the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 and we are given that Then, \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}\) It is given that \(\alpha+\beta+\gamma=6\) Substituting \(\alpha+\beta+\gamma=\frac{3\text{k}}{2},\) we get \(\frac{+3\text{k}}{2}=6\) \(+3\text{k}=6\times2\) \(+3\text{k}=12\) \(\text{k}=\frac{12}{+3}\) \(\text{k}=+4\) The value of k is 4 Hence, the correct alternative is (b)
89960
If \(\alpha,\ \beta\) are the zeros of kx\(^{1}\) - 2x + 3k such that \(\alpha+\beta=\alpha\beta,\) then k = ?
1 \(\frac{1}{3}\)
2 \(\frac{-1}{3}\)
3 \(\frac{2}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
\(\frac{2}{3}\) Here, p(x) = x\(^{1}\) - 2x + 3k Comparing the given polynomial with ax\(^{1}\)+ bx + c, we get: a = 1, b = -2 and c = 3k It is given that \(\alpha\) and \(\beta\) are the roots of the polynomial. \(\therefore\alpha+\beta=-\frac{\text{b}}{\text{a}}\) \(\Rightarrow\alpha+\beta=-\Big(\frac{-2}{1}\Big)\) \(\Rightarrow\alpha+\beta=2\dots(\text{i})\) Also, \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\Rightarrow\alpha\beta=\frac{\text{5k}}1{}\) \(\Rightarrow\alpha\beta=\text{3k}\dots(\text{ii})\) Now, \(\alpha+\beta=\alpha\beta\) \(\Rightarrow2=\text{3k}\) [Using (i) and (ii)] \(\Rightarrow\text{k}=\frac{2}{3}\)
POLYNOMIALS
89961
If \(\alpha\) and \(\beta\) are zeros of x\(^{1}\) + 5x + 8, then the value of \((\alpha+\beta)\) is:
1 8
2 5
3 -5
4 -8
Explanation:
-5 \(\text{x}^{2}+{5}\text{x}+{8}\) \(\alpha+\beta=\frac{\text{Coefficient of x}}{\text{Coefficient of }{\text{x}}^{2}}\) \(=\frac{-5}{1} = {-5}\)
\(\text{x}+\frac{3}{\text{x}}\) An expression of the form p(x) = a\(_{\1}\) + a\(_{\1}\)x + a\(_{\1}\)x\(^{1}\) + ... + a\(_{\1}\)x\(^{1}\), where a\(_{\1}\) ? 0, is called a polynomial in x of degree n. Here, a\(_{\1}\), a\(_{\1}\), a\(_{\1}\), ..., a\(_{\1}\) are real numbers and each power of x is a non-negative integer. Hence, \(\text{x}+\frac{3}{\text{x}}\) is not a polynomial.
POLYNOMIALS
89963
If the sum of the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 is 6, then the value of k is:
1 2
2 4
3 -2
4 -4
Explanation:
4 Let \(\alpha,\beta\) be the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 and we are given that Then, \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}\) It is given that \(\alpha+\beta+\gamma=6\) Substituting \(\alpha+\beta+\gamma=\frac{3\text{k}}{2},\) we get \(\frac{+3\text{k}}{2}=6\) \(+3\text{k}=6\times2\) \(+3\text{k}=12\) \(\text{k}=\frac{12}{+3}\) \(\text{k}=+4\) The value of k is 4 Hence, the correct alternative is (b)
89960
If \(\alpha,\ \beta\) are the zeros of kx\(^{1}\) - 2x + 3k such that \(\alpha+\beta=\alpha\beta,\) then k = ?
1 \(\frac{1}{3}\)
2 \(\frac{-1}{3}\)
3 \(\frac{2}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
\(\frac{2}{3}\) Here, p(x) = x\(^{1}\) - 2x + 3k Comparing the given polynomial with ax\(^{1}\)+ bx + c, we get: a = 1, b = -2 and c = 3k It is given that \(\alpha\) and \(\beta\) are the roots of the polynomial. \(\therefore\alpha+\beta=-\frac{\text{b}}{\text{a}}\) \(\Rightarrow\alpha+\beta=-\Big(\frac{-2}{1}\Big)\) \(\Rightarrow\alpha+\beta=2\dots(\text{i})\) Also, \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\Rightarrow\alpha\beta=\frac{\text{5k}}1{}\) \(\Rightarrow\alpha\beta=\text{3k}\dots(\text{ii})\) Now, \(\alpha+\beta=\alpha\beta\) \(\Rightarrow2=\text{3k}\) [Using (i) and (ii)] \(\Rightarrow\text{k}=\frac{2}{3}\)
POLYNOMIALS
89961
If \(\alpha\) and \(\beta\) are zeros of x\(^{1}\) + 5x + 8, then the value of \((\alpha+\beta)\) is:
1 8
2 5
3 -5
4 -8
Explanation:
-5 \(\text{x}^{2}+{5}\text{x}+{8}\) \(\alpha+\beta=\frac{\text{Coefficient of x}}{\text{Coefficient of }{\text{x}}^{2}}\) \(=\frac{-5}{1} = {-5}\)
\(\text{x}+\frac{3}{\text{x}}\) An expression of the form p(x) = a\(_{\1}\) + a\(_{\1}\)x + a\(_{\1}\)x\(^{1}\) + ... + a\(_{\1}\)x\(^{1}\), where a\(_{\1}\) ? 0, is called a polynomial in x of degree n. Here, a\(_{\1}\), a\(_{\1}\), a\(_{\1}\), ..., a\(_{\1}\) are real numbers and each power of x is a non-negative integer. Hence, \(\text{x}+\frac{3}{\text{x}}\) is not a polynomial.
POLYNOMIALS
89963
If the sum of the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 is 6, then the value of k is:
1 2
2 4
3 -2
4 -4
Explanation:
4 Let \(\alpha,\beta\) be the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 and we are given that Then, \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}\) It is given that \(\alpha+\beta+\gamma=6\) Substituting \(\alpha+\beta+\gamma=\frac{3\text{k}}{2},\) we get \(\frac{+3\text{k}}{2}=6\) \(+3\text{k}=6\times2\) \(+3\text{k}=12\) \(\text{k}=\frac{12}{+3}\) \(\text{k}=+4\) The value of k is 4 Hence, the correct alternative is (b)
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POLYNOMIALS
89960
If \(\alpha,\ \beta\) are the zeros of kx\(^{1}\) - 2x + 3k such that \(\alpha+\beta=\alpha\beta,\) then k = ?
1 \(\frac{1}{3}\)
2 \(\frac{-1}{3}\)
3 \(\frac{2}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
\(\frac{2}{3}\) Here, p(x) = x\(^{1}\) - 2x + 3k Comparing the given polynomial with ax\(^{1}\)+ bx + c, we get: a = 1, b = -2 and c = 3k It is given that \(\alpha\) and \(\beta\) are the roots of the polynomial. \(\therefore\alpha+\beta=-\frac{\text{b}}{\text{a}}\) \(\Rightarrow\alpha+\beta=-\Big(\frac{-2}{1}\Big)\) \(\Rightarrow\alpha+\beta=2\dots(\text{i})\) Also, \(\alpha\beta=\frac{\text{c}}{\text{a}}\) \(\Rightarrow\alpha\beta=\frac{\text{5k}}1{}\) \(\Rightarrow\alpha\beta=\text{3k}\dots(\text{ii})\) Now, \(\alpha+\beta=\alpha\beta\) \(\Rightarrow2=\text{3k}\) [Using (i) and (ii)] \(\Rightarrow\text{k}=\frac{2}{3}\)
POLYNOMIALS
89961
If \(\alpha\) and \(\beta\) are zeros of x\(^{1}\) + 5x + 8, then the value of \((\alpha+\beta)\) is:
1 8
2 5
3 -5
4 -8
Explanation:
-5 \(\text{x}^{2}+{5}\text{x}+{8}\) \(\alpha+\beta=\frac{\text{Coefficient of x}}{\text{Coefficient of }{\text{x}}^{2}}\) \(=\frac{-5}{1} = {-5}\)
\(\text{x}+\frac{3}{\text{x}}\) An expression of the form p(x) = a\(_{\1}\) + a\(_{\1}\)x + a\(_{\1}\)x\(^{1}\) + ... + a\(_{\1}\)x\(^{1}\), where a\(_{\1}\) ? 0, is called a polynomial in x of degree n. Here, a\(_{\1}\), a\(_{\1}\), a\(_{\1}\), ..., a\(_{\1}\) are real numbers and each power of x is a non-negative integer. Hence, \(\text{x}+\frac{3}{\text{x}}\) is not a polynomial.
POLYNOMIALS
89963
If the sum of the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 is 6, then the value of k is:
1 2
2 4
3 -2
4 -4
Explanation:
4 Let \(\alpha,\beta\) be the zeros of the polynomial f(x) = 2x\(^{1}\) - 3kx\(^{1}\) + 4x - 5 and we are given that Then, \(\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}\) It is given that \(\alpha+\beta+\gamma=6\) Substituting \(\alpha+\beta+\gamma=\frac{3\text{k}}{2},\) we get \(\frac{+3\text{k}}{2}=6\) \(+3\text{k}=6\times2\) \(+3\text{k}=12\) \(\text{k}=\frac{12}{+3}\) \(\text{k}=+4\) The value of k is 4 Hence, the correct alternative is (b)
