Question

It is given that the difference between the zeros $4x^2-8kx+9$ of 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

The correct option is C $\frac{5}{2}$

$\Rightarrow$ The given quadratic equation is $4x^2-8kx+9=0$

$\Rightarrow$ $a=4,b=-8k,c=9$

$\Rightarrow$ $\alpha -\beta =4$ ------ ( 1 ) [ Given ]

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-8k}{4}$

$\therefore$ $\alpha +\beta =-2k$ ----- ( 2 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}$

$\therefore$ $\alpha \beta =\frac{9}{4}$ ------- ( 3 )

$\Rightarrow$ $(\alpha +\beta {)}^2=(\alpha -\beta {)}^2+4\alpha \beta$

$\Rightarrow$ $(-2k{)}^2=16+4\times \frac{9}{4}$

$\Rightarrow$ $4k^2=16+9$

$\Rightarrow$ $4k^2=25$

$\Rightarrow$ $k^2=\frac{25}{4}$

$\therefore$ $k=\pm \frac{5}{2}$