Question

find the roots of x

${x}^{2} + x - 12$

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Answer 1

$\huge{\boxed{\sf{SOLUTION-}}}$

It is given that,

${x}^{2} + x - 12$

We have to solve it using the process of middle term splitting.

${x}^{2} + (4 - 3)x \: - 12$

${x}^{2} + 4x - 3x - 12$

$x(x + 4) - 3(x + 4)$

$(x + 4)(x </strong><strong>-</strong><strong> 3)$

Now we will equate both the terms with zero.

$(x + 4) = 0 \\ x = ( - 4)$

And,

$(x </strong><strong>-</strong><strong> 3) = 0 \\ x = ( 3)$

Hence,

The roots of x are (-4) and (3)

Answer 2

Answer:

□ROOT OF X=3 AND -4

Step-by-step explanation:

$your \: question \: is \: roots \: of \: x \: in \: this \: eqn \: \\ so \: first \: we \: use \: \\ middle \: term \: splitting \\ {x}^{2} + x - 12 \\ {x}^{2} + 4x - 3x - 12 \\ x(x + 4) - 3(x + 4) \\ (x - 3)(x + 4) \\ value \: of \: x = 3 \: and \: - 4 \\ \\ second \: method \\ d = {b}^{2} - 4ac \\ \: \: \: = {1}^{2} + 48 = 49 \\ x = \frac{ - b + - \sqrt{d} }{2a} \\ \: \: \: \: = \frac{ - 1 + 7}{2} \: \: \: \: \: \: \: \: \: \frac{ - 1 - 7}{2} \\ \: \: \: \: = \frac{6}{2} \: \: \: \: \: \: \: \: \: \frac{ - 8}{2} \\ x = 3 \: and \: - 4$