If $a^{2} + b^{2} = 13$ and $a b = 6$ , find $a - b$ .

\pm 1

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

\pm 3

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

\pm 9

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

\pm 5

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

Open in App

Solution

The correct option is A $\pm 1$
$(a - b)^{2} = a^{2} + b^{2} - 2 a b$
$= 13 - 2 (6)$
$= 13 - 12$
$= 1$
$=> (a - b) = \pm 1$

Suggest Corrections

Similar questions

a^{2} + b^{2} = 13

a b = 6

a^{2} - b^{2}

a^{2} + b^{2} = 13

a^{2} + b^{2} = 13

a b = 6

3 {(a + b)}^{2} - 2 {(a - b)}^{2}

a^{2} + b^{2}

a + b = 5

a b = 6

a^{2} + b^{2}

Join BYJU'S Learning Program