The min value of $x^{2} - 9 x + 20 = 0$ is at $x = ?$

\frac{9}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- \frac{9}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{2}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{9}{2}$
$f (x) = x^{2} - 9 x + 20$
The extreme value of $f (x)$ is given at $f^{'} (x) = 0$
$f^{'} (x) = \frac{d}{d x} x^{2} - 9 x + 20 = 0 2 x - 9 = 0 2 x = 9 x = \frac{9}{2}$

Suggest Corrections

Similar questions

\sqrt{1 + (a - 6)^{2}} = \sqrt{29} + \sqrt{36 + (4 - 9)^{2}}

Question 79 (b)

Identify the greater number, in each of the following.
$2^{9} o r 9^{2}$

\sqrt{9^{x}} = \sqrt[3]{9^{2}}

x

x^{2} - 3 x - 5 = 0

Join BYJU'S Learning Program