Question

Solve the following quadratic equation using the perfect square trinomial pattern: 14x2+x+4

Answer 1

To use the perfect square trinomial pattern to solve the quadratic equation 14x^2 + x + 4, we need to rewrite the equation in the form (ax + b)^2.

The general form of a perfect square trinomial is (ax + b)^2 = a^2x^2 + 2abx + b^2.

In the given equation, a = sqrt(14), so we need to rewrite the original equation as:

(√14x + b)^2 = 14x^2 + 2b√14x + b^2

Comparing this to the original equation 14x^2 + x + 4, we can see that 2b√14x is equivalent to x, and b^2 is equivalent to 4.

Therefore, we have the following equations:

2b√14x = x
b^2 = 4

From the first equation, we can solve for b by setting up the following equation and solving it for b:

2b√14 = 1
b√14 = 1/2
b = 1/(2√14)

Substituting this value for b into the second equation to solve for b^2, we have:

(1/(2√14))^2 = 4
1/(4√14) = 4
1 = 16(√14)
√14 = 1/16

Now, substituting the value of b = 1/(2√14) into the perfect square trinomial form, we have:

(√14x + 1/(2√14))^2 = 14x^2 + 2(1/(2√14))√14x + (1/(2√14))^2
= 14x^2 + x + 1/56

Therefore, the solution is:

14x^2 + x + 4 = 14x^2 + x + 1/56

The equation is now in the form of a perfect square trinomial.