Question

Find the value of 'k' so that the zeroes of the quadratic polynomial 3x² - kx + 14 are in the ratio 7:6 .​

Answer 1

Answer:

$\boxed{ \red{ \sf \: k = 13}}$

★ GivEn :-

• the zeroes of the quadratic polynomial 3x² - kx + 14 are in the ratio 7:6

★ To Find :-

• The value of k

Step-by-step explanation:

Let the two zeroes of the quadratic polynomial be 7x and 6x ( x > 0 ; common factor )

Formula of a quadratic polynomial,

$\boxed {\sf \: a {x}^{2} + bx + c = 0 \: ;(a \neq 0)}$

We know that,

$\boxed{ \sf \: Sum \: of \: the \: zeroes = - \dfrac{b}{a}}$

$\implies \sf \: 7x + 6x = - (\dfrac{ - k}{3} )$

$\implies \sf \: 13x = \dfrac{ k}{3}$

$\implies \sf \: x = \dfrac{ k}{3 \times 13}$

$\implies \boxed{ \red{ \sf \: x = \dfrac{ k}{39} }} \: ........1)$

And,

$\boxed{ \sf \: Product \: of \: zeroes = \dfrac{c}{a} }$

$\implies \sf \: 7x \times 6x = \dfrac{14}{3}$

$\implies \sf \: 42 {x}^{2} = \dfrac{14}{3}$

$\implies \sf \: {x}^{2} = \dfrac{14}{3 \times 42}$

$\implies \sf \: {x}^{2} = \dfrac{1}{9}$

$\implies \boxed{ \red{\sf \: x = \dfrac{1}{3} }}$

Putting the value of x in the place of equation no 1) we get,

$\implies \sf \: \dfrac{1}{3} = \dfrac{k}{39}$

$\implies \sf \: 3k = 39$

$\implies \boxed{ \red{ \therefore \: \sf \: k = \cancel\dfrac{39}{3} = 13}}$

Answer 2

$\huge\sf\purple{k=±13}$

$\sf Let \:\alpha\: and\: \beta \:be\: the \:roots \:of\: polynomial \\\\\sf \: 3x² - kx + 14 \\\\\sf</p><p>Then, \:\alpha + \beta = \frac{-(-k)} {3}=\frac{k}{3}\:\:\: (i) \\\\\sf</p><p>and\: \alpha\: x \:\beta = \frac{14}{3} \:\:\:(ii)\\\\\sf </p><p>We \:know\: that\: \alpha:\beta=7:6 \\\\\sf</p><p>6\alpha=7\beta \\\\\sf</p><p>\alpha=\frac{7\beta} {6} \\\\\sf</p><p></p><p>Putting, \:\alpha=\frac{7\beta}{6} in (i), we\:get \\\\\sf</p><p>=\alpha + \beta =\frac{k}{3} \\\\\sf</p><p>=\frac{7\beta} {6} + \beta= \frac{k} {3} \\\\\sf</p><p>=\frac{7\beta+6\beta}{6}=\frac{k} {3} \\\\\sf</p><p>=\frac{13\beta} {6}=\frac{k} {3} \\\\\sf</p><p>=2k=13\beta \:(iii) \\\\\sf</p><p>Now,\: from\: (ii) \\\\\sf</p><p>=\alpha \:x \:\beta = \frac{14}{3} \\\\\sf</p><p>=\frac{7\beta} {6} x \beta=\frac{14}{3} \\\\\sf</p><p>=\frac{7\beta^2}{6} =\frac{14}{3} \\\\\sf</p><p>=\beta^2=4 \\\\\sf</p><p>=\beta=±2 \\\\\sf</p><p>Putting,\: \beta=\:±2\: in \: (iii), we\: get \\\\\sf</p><p>=2k=13\beta \\\\\sf</p><p>=2k=13\:x±2 \\\\\sf</p><p>k=\frac{13\:x±2}{2} \\\\\sf</p><p>k=\frac{13\:x± \cancel{2}}{ \cancel{2}} \\\\\sf</p><p>\huge\boxed{k=±13} \\\\\sf</p><p>$