90019
If \(\alpha\) and \(\beta\) are the zero of 2x\(^{1}\) + 5x − 8, then the value of \((\alpha+\beta)\) is:
1 \(\frac{-5}{2}\)
2 \(\frac{5}{2}\)
3 \(\frac{-9}{2}\)
4 \(\frac{9}{2}\)
Explanation:
\(\frac{-9}{2}\) Let \(\alpha\) and \(\beta\) be the zeros of the 2x\(^{1}\) + 5x - 9 Then, we have \(\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}\)
POLYNOMIALS
90151
If one zero of the quadratic polynomial kx\(^{1}\) + 3x + k is 2, then the value of k is:
1 \(\frac{5}{6}\)
2 \(\frac{-5}{6}\)
3 \(\frac{6}{5}\)
4 \(\frac{-6}{5}\)
Explanation:
\(\frac{-6}{5}\) Since 2 is a zero of f(x) = kx\(^{1}\) + 3x + k, we have f(2) = 0 ? k(2)\(^{1}\) + 3(2) + k = 0 ? 4k + 6 + k = 0 ? 5k + 6 = 0 ? 5k = -6 \(\Rightarrow\text{k}=-\frac{6}{5}\)
POLYNOMIALS
90070
If two of the zeroes of a cubic polynomial ax\(^{1}\)+ bx\(^{1}\)+ cx + d are zero, then the third zero is:
90019
If \(\alpha\) and \(\beta\) are the zero of 2x\(^{1}\) + 5x − 8, then the value of \((\alpha+\beta)\) is:
1 \(\frac{-5}{2}\)
2 \(\frac{5}{2}\)
3 \(\frac{-9}{2}\)
4 \(\frac{9}{2}\)
Explanation:
\(\frac{-9}{2}\) Let \(\alpha\) and \(\beta\) be the zeros of the 2x\(^{1}\) + 5x - 9 Then, we have \(\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}\)
POLYNOMIALS
90151
If one zero of the quadratic polynomial kx\(^{1}\) + 3x + k is 2, then the value of k is:
1 \(\frac{5}{6}\)
2 \(\frac{-5}{6}\)
3 \(\frac{6}{5}\)
4 \(\frac{-6}{5}\)
Explanation:
\(\frac{-6}{5}\) Since 2 is a zero of f(x) = kx\(^{1}\) + 3x + k, we have f(2) = 0 ? k(2)\(^{1}\) + 3(2) + k = 0 ? 4k + 6 + k = 0 ? 5k + 6 = 0 ? 5k = -6 \(\Rightarrow\text{k}=-\frac{6}{5}\)
POLYNOMIALS
90070
If two of the zeroes of a cubic polynomial ax\(^{1}\)+ bx\(^{1}\)+ cx + d are zero, then the third zero is:
90019
If \(\alpha\) and \(\beta\) are the zero of 2x\(^{1}\) + 5x − 8, then the value of \((\alpha+\beta)\) is:
1 \(\frac{-5}{2}\)
2 \(\frac{5}{2}\)
3 \(\frac{-9}{2}\)
4 \(\frac{9}{2}\)
Explanation:
\(\frac{-9}{2}\) Let \(\alpha\) and \(\beta\) be the zeros of the 2x\(^{1}\) + 5x - 9 Then, we have \(\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}\)
POLYNOMIALS
90151
If one zero of the quadratic polynomial kx\(^{1}\) + 3x + k is 2, then the value of k is:
1 \(\frac{5}{6}\)
2 \(\frac{-5}{6}\)
3 \(\frac{6}{5}\)
4 \(\frac{-6}{5}\)
Explanation:
\(\frac{-6}{5}\) Since 2 is a zero of f(x) = kx\(^{1}\) + 3x + k, we have f(2) = 0 ? k(2)\(^{1}\) + 3(2) + k = 0 ? 4k + 6 + k = 0 ? 5k + 6 = 0 ? 5k = -6 \(\Rightarrow\text{k}=-\frac{6}{5}\)
POLYNOMIALS
90070
If two of the zeroes of a cubic polynomial ax\(^{1}\)+ bx\(^{1}\)+ cx + d are zero, then the third zero is:
90019
If \(\alpha\) and \(\beta\) are the zero of 2x\(^{1}\) + 5x − 8, then the value of \((\alpha+\beta)\) is:
1 \(\frac{-5}{2}\)
2 \(\frac{5}{2}\)
3 \(\frac{-9}{2}\)
4 \(\frac{9}{2}\)
Explanation:
\(\frac{-9}{2}\) Let \(\alpha\) and \(\beta\) be the zeros of the 2x\(^{1}\) + 5x - 9 Then, we have \(\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}\)
POLYNOMIALS
90151
If one zero of the quadratic polynomial kx\(^{1}\) + 3x + k is 2, then the value of k is:
1 \(\frac{5}{6}\)
2 \(\frac{-5}{6}\)
3 \(\frac{6}{5}\)
4 \(\frac{-6}{5}\)
Explanation:
\(\frac{-6}{5}\) Since 2 is a zero of f(x) = kx\(^{1}\) + 3x + k, we have f(2) = 0 ? k(2)\(^{1}\) + 3(2) + k = 0 ? 4k + 6 + k = 0 ? 5k + 6 = 0 ? 5k = -6 \(\Rightarrow\text{k}=-\frac{6}{5}\)
POLYNOMIALS
90070
If two of the zeroes of a cubic polynomial ax\(^{1}\)+ bx\(^{1}\)+ cx + d are zero, then the third zero is:
90019
If \(\alpha\) and \(\beta\) are the zero of 2x\(^{1}\) + 5x − 8, then the value of \((\alpha+\beta)\) is:
1 \(\frac{-5}{2}\)
2 \(\frac{5}{2}\)
3 \(\frac{-9}{2}\)
4 \(\frac{9}{2}\)
Explanation:
\(\frac{-9}{2}\) Let \(\alpha\) and \(\beta\) be the zeros of the 2x\(^{1}\) + 5x - 9 Then, we have \(\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}\)
POLYNOMIALS
90151
If one zero of the quadratic polynomial kx\(^{1}\) + 3x + k is 2, then the value of k is:
1 \(\frac{5}{6}\)
2 \(\frac{-5}{6}\)
3 \(\frac{6}{5}\)
4 \(\frac{-6}{5}\)
Explanation:
\(\frac{-6}{5}\) Since 2 is a zero of f(x) = kx\(^{1}\) + 3x + k, we have f(2) = 0 ? k(2)\(^{1}\) + 3(2) + k = 0 ? 4k + 6 + k = 0 ? 5k + 6 = 0 ? 5k = -6 \(\Rightarrow\text{k}=-\frac{6}{5}\)
POLYNOMIALS
90070
If two of the zeroes of a cubic polynomial ax\(^{1}\)+ bx\(^{1}\)+ cx + d are zero, then the third zero is: