Find the value of $^{'} a^{'}$ in $4 x^{2} + a x + 9 = (2 x + 3)^{2}$ .

a = 34

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a = 11

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a = 12

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a = 33

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Solution

The correct option is B $a = 12$
$4 x^{2} + a x + 9 = (2 x + 3)^{2}$
$=> 4 x^{2} + a x + 9 = 4 x^{2} + 2 (2 x) (3) + 9$
$=> a x = 2 (2 x) (3)$
$=> a x = 12 x$
$=> a = 12$

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Similar questions

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

^{'} a^{'}

4 x^{2} + a x + 9 = {(2 x + 3)}^{2}

Q. Find the value of

3^{2} \times 3^{4} \times 3^{3} = ?

If $Δ ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣$ and $A_{i j}$ is cofactor of $a_{i j}$ , then value of $Δ$ is given by

a) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$
b) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$
c) $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$
d) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$

Q. Find the remainder when

32^{33^{34}}

is divided by 11.

Q. Solve

\frac{(4 x + 1)^{2} + (2 x + 3)^{2}}{4 x^{2} + 12 x + 9} = \frac{61}{36}

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Find the value of ′a′ in 4x2+ax+9=(2x+3)2.

The correct option is B a=12 4x2+ax+9=(2x+3)2=>4x2+ax+9=4x2+2(2x)(3)+9=>ax=2(2x)(3)=>ax=12x=>a=12

Find the value of $^{'} a^{'}$ in $4 x^{2} + a x + 9 = (2 x + 3)^{2}$ .

The correct option is B $a = 12$
$4 x^{2} + a x + 9 = (2 x + 3)^{2}$
$=> 4 x^{2} + a x + 9 = 4 x^{2} + 2 (2 x) (3) + 9$
$=> a x = 2 (2 x) (3)$
$=> a x = 12 x$
$=> a = 12$