In the following equation given below, find the value of $^{'} a^{'}$ :
$9 x^{2} + (7 a - 5) x + 25 = {(3 x + 5)}^{2}$

a = 4

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

a = - 14

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

a = - 12

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

a = - 2

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

Open in App

Solution

The correct option is C $a = - 12$
$9 x^{2} + (7 a - 5) x + 25 = {(3 x + 5)}^{2}$
$=> 9 x^{2} + (7 a - 5) x + 25 = 9 x^{2} + 2 (3 x) (5) + 25$
$=> (7 a - 5) x = 30 x$
$=> 7 a = 35$
$=> a = 5$

Suggest Corrections

Similar questions

a

9 x^{2} + (7 a - 5) x + 25 = (3 x + 5)^{2}

\frac{5 + 3 \sqrt{2}}{5 - 3 \sqrt{2}} = a + b \sqrt{2}

3 x, + 5 y = 12, 5 x + 3 y = 4

\frac{9 x^{2}}{3 x - 2}

= A + \frac{4}{3 x - 2}

A

x

25 [(x - 5)^{2} + (y - 3)^{2}] = (3 x - 4 y + 1)^{2}

Join BYJU'S Learning Program