Find the value of μ for which one root of the quadratic equation μx2−14x+8 = 0 is 6 times the other.

# Find the value of μ for which one root of the quadratic equation μx2−14x+8 = 0 is 6 times the other.

1. A
5
2. B
4
3. C
3
4. D
2

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### Solution:

Given: The quadratic equation is μx2−14x+8 = 0
One root of the equation given is 6 times the other.
We need to find the value of μ
Let the roots of the given quadratic equation be α and β respectively.
According to the question,
α is 6 times of β
⇒α=6×β
Sum of the roots of the equation in general form ax2+bx+c=0 is $-\frac{b}{a}$
Comparing the given equation μx2−14x+8 = 0 with the general form, we get
a=μ
b=−14
c=8
α+β$=-\frac{b}{a}$  = $-\left(-\frac{14}{\mu }\right)$

Given that α=6×β
Substituting α=6×β in Equation (1)
We get,
$6\beta +\beta =\frac{14}{\mu }$
$⇒7\beta =\frac{14}{\mu }$

Product of the roots of the equation in general form ax2+bx+c=0 is ca

Substituting α=6×β in Equation (3)
We get,
$6\beta ×\beta =\frac{8}{\mu }$

Substituting Equation (2) in Equation (4) , we get

$⇒6×\frac{4}{{\mu }^{2}}=\frac{8}{\mu }$
$⇒\mu =3$
Therefore, the value of μ is 3.

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