Question

Find the value of k in the polynomial
10x2 + kx + 12, such that the zeroes of the
polynomial will have a ratio of 8:15.​

Answer 1

Given

• f(x) = 10x² + kx + 12
• Ratio of Zeros is 8:15

To Find

• Value of k

Solution

f(x) = 10x² + kx + 12

a=10, b=k, c= 12

$let \: \alpha \: and \: \beta \: be \: the \: zeroes \: of \: f(x).$

Let

$\alpha = 8 x\\ \beta = 15x$

$Sum \: of \: Zeros \: ( \alpha + \beta )=\dfrac{-b}{a}$

$\: 8x + 15x=\dfrac{-k}{10}$

$\: 23x=\dfrac{-k}{10} \: \: \: \: \: - - - - (1)$

Now

$Product \: of \: Zeros \: ( \alpha \beta ) = \dfrac{c}{a}$

$8x \times 15x= \dfrac{12}{10}$

$120 {x}^{2} = \dfrac{12}{10}$

${x}^{2} = \dfrac{12}{10 \times 120}$

${x}^{2} = \dfrac{1}{10 \times 10}$

${x}^{2} = \dfrac{1}{100}$

${x}= \dfrac{1}{10}$

• Putting x in Equation 1

$23 \times \dfrac{1}{10} =\dfrac{-k}{10}$

$\dfrac{-k}{10} \times 10 = 23$

$- k = 23$

$k = - 23$

• Hance the Value of k is -23