The number of real solutions of the equation:
$e^{x} = x^{2} + 14 x + 49$ , is

0

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

1

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

2

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

none of these

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

Open in App

Solution

The correct option is C $2$

$x^{2} + 14 x + 49 = {(x + 7)}^{2}$

$e^{x} = {(x + 7)}^{2}$

Curves of $e^{x}$ and ${(x + 7)}^{2}$ intersect at two points

Suggest Corrections

Similar questions

| x^{2} | - 3 | x | + 2 = 0

x^{2} - x + 3 + {(\frac{9}{10})}^{x} = 0

e^{x} + x = 0

2 | x |^{2} - 5 | x | + 2 = 0

k, b

x^{2} + b x - 2 + k (x^{2} + 3 x + 2) = 0

Join BYJU'S Learning Program