Find the value of ' $a$ ' in $9 x^{2} + (7 a - 5) x + 25 = (3 x + 5)^{2}$ .

a = 1

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a = 5

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a = - 3

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a = - 7

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Solution

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

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Q. In the following equation given below, find the value of

^{'} a^{'}

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

Write 'T' for true and 'F' for false
(i) In $- x$ , the constant term is $- 1$ .
(ii) The coefficient of $x$ in $x^{2} - 3 x + 5$ is $3$ .
(iii) $(5 x - 7) - (3 x - 5) = 2 x - 12$ .
(iv) $(3 x + 5 y) (3 x - 5 y) = (9 x^{2} - 25 y^{2})$ .
(v) If $a = 2$ and $b = 12$ , then the value of $a b (a^{2} + b^{2})$ is $414$ .