Question

What are the roots of the equation X2 – 12x = –27?

Answer 1

SOLUTION

TO DETERMINE

The roots of the equation

$\sf{ {x}^{2} - 12x = - 27}$

EVALUATION

Here the given equation is

$\sf{ {x}^{2} - 12x = - 27}$

We find the roots as below

$\sf{ {x}^{2} - 12x = - 27}$

$\sf{ \implies \: {x}^{2} - 12x + 27 = 0}$

$\sf{ \implies \: {x}^{2} - (3 + 9)x + 27 = 0}$

$\sf{ \implies \: {x}^{2} - 3x - 9x + 27 = 0}$

$\sf{ \implies \: x(x - 3) - 9(x - 3) = 0}$

$\sf{ \implies \: (x - 3) (x - 9) = 0}$

Now x - 3 = 0 gives x = 3

x - 9 = 0 gives x = 9

FINAL ANSWER

Hence the required roots are 3 & 9

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Answer 2

Answer:

The roots of the equation $x^2-12x=-27$ are $9$ and $3$.

Step-by-step explanation:

Step 1 of 2

Consider the quadratic equation as follows:

$x^2-12x=-27$

Rewrite the equation as follows:

$x^2-12x+27=0$

Using the middle term splitting method,

We need to find two integers $a$ and $b$ such that $a+b=-12$ and $ab=27$.

Look into the pairs of factors of $27$.

$-1$ and $-27$, $-3$ and $-9$, are the pairs of factors of $27$.

Notice that the pair $-3$ and $-9$ gives sum as $-12$ and the product as $27$.

Step 2 of 2

Factorising the given equation as follows:

⇒ $x^2-9x-3x+27=0$

⇒ $x(x-9)-3(x-9)=0$

Take the term $(x-9)$ common out.

⇒ $(x-9)(x-3)=0$

⇒ $x=9$ and $x=3$

Therefore, the roots of the equation $x^2-12x=-27$ are $9$ and $3$.

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