Question

if one zero of the equation 3x²-8x+2k+1 is seven times the other .find the zeroes and value of k.​

Answer 1

Answer:

$\large k=\dfrac{2}{3}$

Step-by-step explanation:

$\large \text{Given p(x)=3x^2-8x+2k+1 }\\\\\\\large \text{Also given one zero seven times the other }\\\\\\\large \text{Let one zero \alpha \ and \ other \ zero \ will \ be \ 7 \alpha }$

$\large \text{We know product of two zero \alpha \times \beta =\dfrac{c}{a} }$

where c is coefficient of constant term and a is coefficient of  $x^2$

Now putting values here we get

$\large \alpha \times \beta =\dfrac{2k+1}{3}\\\\\\\large \text{Here \beta=7\alpha }\\\\\\\large 3\times7(\alpha)^2=2k+1\\\\\\\large 21(\alpha)^2=2k+1 \ ....(i)$

$\large \text{We also know sum of two zero \alpha+\beta =\dfrac{-b}{a} }$

where b is coefficient of x term and a is coefficient of  $x^2$

putting values here

$\large \alpha+7\alpha=\dfrac{8}{3}\\\\\\\large 8\alpha=\dfrac{8}{3}\\\\\\\large 1\alpha=\dfrac{1}{3}\\\\\\\large \alpha=\dfrac{1}{3}\\\\\\\large \text{Now putting in (i) }$

$\large 21(\dfrac{1}{3} )^2=2k+1 \\\\\\\large \dfrac{7}{3}=2k+1\\\\\\\large \dfrac{7}{3}-1=2k\\\\\\\large k=\dfrac{2}{3}$

Thus we get k = 2 / 3