The sum of zeroes of the polynomial f(x) = 2x³- 3kx² + 4x – 5 is 6, then the value of k is?
ANSWER:
Given:
f(x) = 2x³- 3kx² + 4x – 5
Sum of zeroes = 6
To Find:
Value of k
Solution:
[tex]\text{We are given that,}\\\:\longrightarrow f(x)=2x^3-3kx^2+4x-5\\\\\text{We know that,}\\\\\text{For a cubic polynomial, $p(x)=ax^2+bx^2+cx+d$,}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{\text{Coefficient of x$^2$}}{\text{Coefficient of x$^3$}}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{b}{a}\\\\\text{So,}\\\\\text{Here, a = 2 and b = -3k. So,}\\\\:\implies\text{Sum of zeroes}=-\dfrac{-3k}{2}\\\\:\implies\text{Sum of zeroes}=\dfrac{3k}{2}[/tex]
[tex]\text{But, we are given that,}\\\\:\longrightarrow\text{Sum of zeroes}=6\\\\\text{So,}\\\\:\implies\dfrac{3k}{2}=6\\\\\text{On cross-multiplying,}\\\\:\implies3k=6\times2\\\\:\implies3k=12\\\\\text{On transposing 3 to RHS,}\\\\:\implies k=\dfrac{12\!\!\!\!\!/^{\:\:4}}{3\!\!\!/}\\\\\bf{:\implies k=4}\\\\\text{\bf{Hence, value of ‘k’ is 4(Option C).}}[/tex]
Formula Used:
[tex]\text{For a cubic polynomial, $p(x)=ax^2+bx^2+cx+d$,}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{\text{Coefficient of x$^2$}}{\text{Coefficient of x$^3$}}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{b}{a}[/tex]
Answer:
4 is your answer hope it helps you
Step-by-step explanation:
Brainliest please
QUESTION:
ANSWER:
Given:
To Find:
Solution:
[tex]\text{We are given that,}\\\:\longrightarrow f(x)=2x^3-3kx^2+4x-5\\\\\text{We know that,}\\\\\text{For a cubic polynomial, $p(x)=ax^2+bx^2+cx+d$,}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{\text{Coefficient of x$^2$}}{\text{Coefficient of x$^3$}}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{b}{a}\\\\\text{So,}\\\\\text{Here, a = 2 and b = -3k. So,}\\\\:\implies\text{Sum of zeroes}=-\dfrac{-3k}{2}\\\\:\implies\text{Sum of zeroes}=\dfrac{3k}{2}[/tex]
[tex]\text{But, we are given that,}\\\\:\longrightarrow\text{Sum of zeroes}=6\\\\\text{So,}\\\\:\implies\dfrac{3k}{2}=6\\\\\text{On cross-multiplying,}\\\\:\implies3k=6\times2\\\\:\implies3k=12\\\\\text{On transposing 3 to RHS,}\\\\:\implies k=\dfrac{12\!\!\!\!\!/^{\:\:4}}{3\!\!\!/}\\\\\bf{:\implies k=4}\\\\\text{\bf{Hence, value of ‘k’ is 4(Option C).}}[/tex]
Formula Used:
[tex]\text{For a cubic polynomial, $p(x)=ax^2+bx^2+cx+d$,}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{\text{Coefficient of x$^2$}}{\text{Coefficient of x$^3$}}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{b}{a}[/tex]