Question

If $\alpha ,\beta$ are the two roots of the quadratic equation $4x^2-26x+34=0$ such that $\alpha >\beta ,$ then $\alpha =$

• A. $\frac{13-\sqrt{33}}{4}$
• B. $\frac{13+\sqrt{33}}{4}$
• C. $\frac{-13+\sqrt{33}}{4}$

Answer 1

The correct option is B $\frac{13+\sqrt{33}}{4}$

Given: $\alpha ,\beta$ are the roots of the quadratic equation $4x^2-26x+34=0$ such that $\alpha >\beta$
Now, the equation can be simplified as:
$2x^2-13x+17=0$

Now, comparing the equation with $a{x}^2+bx+c=0,$ we get:
$a=2,b=-13,c=17$
Now, roots of the equation $a{x}^2+bx+c=0$ are given by:
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Thus, substituting the values of $a,b\&c$ in the quadratic formula,
we get the roots as
$x=\frac{13\pm \sqrt{{13}^2-4\cdot 2\cdot 17}}{4}$
Since, $\alpha >\beta$
$\Rightarrow \alpha =\frac{13+\sqrt{169-136}}{4}=\frac{13+\sqrt{33}}{4}$
Hence, $\alpha =\frac{13+\sqrt{33}}{4}$