Question

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Answer 1

$\bold{\boxed{HEY!!!}}$

Here is your answer :-

$x + 6x = 7 \\ \\ square \: term \: = {( \frac{1}{2} \times coefficient \: of \: x) }^{2} \\ \\ = {( \frac{1}{2} \times 6) }^{2} \\ \\ = {3 }^{2} \\ \\ = 9 \\ \\ so \: adding \: 9 \: on \: the \: both \: sides \: \\ \\ {x}^{2} + 6x + 9 = 7 + 9 \\ \\ {x}^{2} + 2(3)x + {3}^{2} = 16 \\ \\ {(x + 3)}^{2} = 16 \\ \\ taking \: square \: root \: on \: both \: sides \\ \\ x + 3 = + - 4 \\ \\ x = 4 - 3 \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: x = - 4 - 3 \\ \\ x = 1 \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: x = - 7$


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Answer 2

Hi!!!

x² +2 (x) (3) + 9 -9 -7 = 0

x ² + 2 ( x ) ( 3 ) + 9 -16= 0

(x + 3 )² = 16

x +3 = 4. or. x +3 = -4

x = 4 - 3. or x = -4 -3

x = -1. or. x = -7


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