Question

It is given that the difference between the zeroes of 4x² − 8kx + 9 is 4 and k > 0. Then, k = ?

(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{2}$
(d) $\frac{7}{2}$

Answer 1

(c) $\frac{5}{2}$
Let the zeroes of the polynomial be $\mathrm{\alpha }\mathrm{}\mathrm{and}\mathrm{\alpha }+4$.
Here, p$\left(x\right)\mathrm{}=4x^2-8kx+9$
Comparing the given polynomial with $a{x}^2+bx+c$, we get:
a = 4, b = −8k and c = 9
Now, sum of the roots$=-\frac{b}{a}$
$$\begin{gathered} =>\mathrm{\alpha }+\mathrm{\alpha }+4=\frac{-(-8k)}{4} \\ =>2\mathrm{\alpha }+4=2k \\ =>\mathrm{\alpha }+2=k \\ =>\mathrm{\alpha }=(k-2)...\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{roots},\mathrm{}\mathrm{\alpha \beta }=\frac{c}{a} \\ =>\mathrm{\alpha }(\mathrm{\alpha }+4)=\frac{9}{4} \\ =>(k-2)(k-2+4)=\frac{9}{4} \\ =>\left(k-2\right)\left(k+2\right)=\frac{9}{4} \\ =>k^2-4=\frac{9}{4} \\ =>4k^2-16=9 \\ =>4k^2=25 \\ =>k^2=\frac{25}{4} \\ =>k=\frac{5}{2}(\because k>0) \end{gathered}$$