Please help me with this
1 answer:
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Let's assume,
Hence, the given equation can be written as:
y² - y - 6 = 0
Next step is to factor the trinomial. For that, break down the constant -6 into two multiples so that their addition will result the coefficient of y =-1.
So, 6 = -3* 2
And sum of -3 and 2 will give -1.
Therefore, we can replace -y with -3y + 2y. So,
y² - 3y + 2y - 6 =0
(y² - 3y) + (2y - 6) =0 Make the group of terms.
y(y - 3)+ 2(y -3) =0 Take out the common factor from each group.
(y - 3 ) (y + 2) = 0 Take out the common factor (y - 3).
So, y - 3 = 0 and y + 2 =0 Equate both factor equal to 0.
Hence, y = 3 and -2.
So,
Taking ln to each sides.
So, x = ln 3
But ln (-2) is not defined.
So, x = ln (3) is the real solution of the given equation.