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7x^2- 27x - 4
This equation is a quadratic equation of type Ax^2 + Bx + C
Now, here A = 7 ; B = -27 & C = -4
A*C = -28
B^2 = 729
Now, -27 = -28 + 1
Thus, the equation can be rewritten as 7x^2 - 28x + x - 4
or, 7x(x-4) +1(x-4)
So, none of the given options are correct.
We have to determine which of the given options is equivalent to 7x^2- 27x - 4.
7x^2 - 27x - 4 = 0
=> 7x^2 - 28x + x - 4 = 0
=> 7x(x - 4) + 1(x - 4) = 0
=> (7x + 1)(x - 4)
The options are:
A) (x + 4)(7x - 1) = 7x^2 + 28x - x - 4 = 7x^2 + 27x - 4
B) (x - 2)(7x + 2) = 7x^2 - 14x + 2x - 4 = 7x^2 - 12x - 4
C) (x + 2)(7x - 2) = 7x^2 + 14x - 2x - 4 = 7x^2 + 12x - 4
E) x(7x - 27 - 4) = 7x^2 - 27x - 4x = 7x^2 - 31x
None of the options given are equivalent to the given expression. The factorized form of 7x^2- 27x - 4 = (7x + 1)(x - 4)
All we have to do is to determine the roots of the quadratic, knowing the fact that the quadratic is the result of the product of two linear factors: (x-x1)(x-x2), where x1 and x2 are the roots of the quadratic.
We'll determine the roots applying quadratic formula:
x1 = [27+sqrt(729+112)]/14
x2 = -1/7
We'll write the quadratic:
(x-x1)(x-x2)=(x-4)(x+1/7) = (x-4)(7x+1)
The correct answer is: (x - 4)(7x + 1).
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