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

Question

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

in progress 0
Piper 6 days 2021-11-22T00:46:25+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-11-22T00:47:34+00:00

    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}}

    0
    2021-11-22T00:48:08+00:00

     \huge\sf\purple{k=±13}

     \sf Let \:\alpha\: and\: \beta \:be\: the  \:roots \:of\: polynomial \\\\\sf \: 3x² - kx + 14 \\\\\sfThen, \:\alpha + \beta = \frac{-(-k)} {3}=\frac{k}{3}\:\:\: (i)  \\\\\sfand\: \alpha\: x \:\beta = \frac{14}{3} \:\:\:(ii)\\\\\sf We \:know\: that\: \alpha:\beta=7:6 \\\\\sf6\alpha=7\beta \\\\\sf\alpha=\frac{7\beta} {6}  \\\\\sfPutting, \:\alpha=\frac{7\beta}{6} in (i), we\:get \\\\\sf=\alpha + \beta =\frac{k}{3} \\\\\sf=\frac{7\beta} {6} + \beta= \frac{k} {3} \\\\\sf=\frac{7\beta+6\beta}{6}=\frac{k} {3} \\\\\sf=\frac{13\beta} {6}=\frac{k} {3} \\\\\sf=2k=13\beta \:(iii) \\\\\sfNow,\: from\: (ii)  \\\\\sf=\alpha \:x \:\beta = \frac{14}{3} \\\\\sf=\frac{7\beta} {6} x \beta=\frac{14}{3} \\\\\sf=\frac{7\beta^2}{6} =\frac{14}{3} \\\\\sf=\beta^2=4 \\\\\sf=\beta=±2 \\\\\sfPutting,\: \beta=\:±2\: in \: (iii), we\: get  \\\\\sf=2k=13\beta \\\\\sf=2k=13\:x±2 \\\\\sfk=\frac{13\:x±2}{2} \\\\\sfk=\frac{13\:x± \cancel{2}}{ \cancel{2}} \\\\\sf\huge\boxed{k=±13} \\\\\sf

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )