Question

Find the value of k, ​ ​for which one root of the quadratic equation kx2​ - 14x+ 8 = 0, is six times the other.

Answer 1

${\mathscr{\red{\bold{Heya! }}}}$

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${\underline{\bold{Given:-}}}$

$\alpha + \beta = - \frac{ - b}{a} \\ 6 \beta + \beta = \frac{ - ( - 14)}{k} \\ 7 \beta = \frac{14}{k} \\ \beta = \frac{2}{k}$

from eq(1),

$\alpha \beta = \frac{c}{a} \\ 6 \beta \times \beta = \frac{8}{k } \\ 6 { \beta }^{2} = \frac{8}{k} \\ 6 \times \frac{ {2}^{2} }{ {k}^{2} } = \frac{8}{k} \\ 6 \times \frac{4}{ {k}^{2} } = \frac{8}{k} \\ \frac{24}{ {k}^{2} } = \frac{8}{k} \\ \frac{24}{ {k}^{2} } \times k = 8 \\ \frac{24}{k} = 8 \\ k = \frac{24}{8} \\ k = 3$

hence we get the value of k as,

${\boxed{\blue{\bold{3}}}}$

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Answer 2

The value of k is 3.

Step-by-step explanation:

Given : One root of the quadratic equation $kx^2-14x+ 8 = 0$, is six times the other.

To find : The value of k ?

Solution :

The roots of quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$ satisfy

Sum of roots  $\alpha + \beta =-\frac{b}{a}$

Product of roots $\alpha \beta = \frac{c}{a}$

One roots is 6 time other i.e. $\alpha=6\beta$

Here, a=k, b=-14 and c=8

Now, $6\beta+ \beta=-\frac{-14}{k}$

$7\beta=\frac{14}{k}$

$\beta = \frac{2}{k}$ ....(1)

and $6 \beta \times \beta=\frac{8}{k}$

$6 {\beta }^{2} = \frac{8}{k}$

$6 \times \frac{ {2}^{2} }{ {k}^{2} } = \frac{8}{k}$

$6 \times \frac{4}{ {k}^{2} } = \frac{8}{k}$

$\frac{24}{ {k}^{2} } = \frac{8}{k}$

$\frac{24}{ {k}^{2} } \times k = 8$

$\frac{24}{k} = 8$

$k=\frac{24}{8}$

$k = 3$

Therefore, the value of k is 3.

#Learn more

For what values of k the quadratic equation kx2-14x+8=0 is 2

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