How do you solve #27 x + 4 = 40 x ^2# using the quadratic formula?

Question

Abigail Allen · Accepted Answer

#4/5 and (-1/8)# Solve the quadratic equation: #y = 40x^2 - 27x - 4 = 0# Use the improved quadratic formula (Socratic Search) #D = d^2 = b^2 - 4ac = 729 + 640 = 1369# --> #d = +- 37# There are 2 real roots: #x = -b/(2a) +- d/(2a) = 27/80 +- 37/80 = (27 +- 37)/80# #x1 = 64/80 = 4/5# #x2 = - 10/80 = - 1/8#

Jace Anderson · Answer

undefined To solve the equation $27x + 4 = 40x^2$ using the quadratic formula, follow these steps:

1. Rewrite the equation in standard quadratic form: $40x^2 - 27x - 4 = 0$.
2. Identify the coefficients: $a = 40$, $b = -27$, and $c = -4$.
3. Substitute these values into the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
4. Plug in the values: $x = \frac{{-(-27) \pm \sqrt{{(-27)^2 - 4 \cdot 40 \cdot (-4)}}}}{{2 \cdot 40}}$.
5. Simplify: $x = \frac{{27 \pm \sqrt{{729 + 640}}}}{{80}}$.
6. Further simplify: $x = \frac{{27 \pm \sqrt{{1369}}}}{{80}}$.
7. Calculate the square root: $\sqrt{{1369}} = 37$.
8. Continue simplifying: $x = \frac{{27 \pm 37}}{{80}}$.
9. Find the two possible solutions: $x_1 = \frac{{27 + 37}}{{80}}$ and $x_2 = \frac{{27 - 37}}{{80}}$.
10. Calculate each solution: $x_1 = \frac{{64}}{{80}}$ and $x_2 = \frac{{-10}}{{80}}$.
11. Simplify each solution: $x_1 = \frac{{4}}{{5}}$ and $x_2 = -\frac{{1}}{{8}}$.

Therefore, the solutions to the equation are $x = \frac{{4}}{{5}}$ and $x = -\frac{{1}}{{8}}$.

Savannah Anderson · Answer

undefined To solve the quadratic equation $27x + 4 = 40x^2$ using the quadratic formula, which is given by $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$, we first rewrite the given equation in the standard quadratic form $ax^2 + bx + c = 0$.

Given equation: $27x + 4 = 40x^2$

Rearrange it to bring all terms to one side: $40x^2 - 27x - 4 = 0$

Now, identify the coefficients:
$a = 40$, $b = -27$, and $c = -4$

Substitute these values into the quadratic formula and solve for $x$:
$$x = \frac{{-(-27) \pm \sqrt{{(-27)^2 - 4 \cdot 40 \cdot (-4)}}}}{{2 \cdot 40}}$$

$$x = \frac{{27 \pm \sqrt{{729 + 640}}}}{{80}}$$

$$x = \frac{{27 \pm \sqrt{{1369}}}}{{80}}$$

$$x = \frac{{27 \pm 37}}{{80}}$$

This yields two solutions:
$$x_1 = \frac{{27 + 37}}{{80}} = \frac{{64}}{{80}} = \frac{{4}}{{5}}$$

$$x_2 = \frac{{27 - 37}}{{80}} = \frac{{-10}}{{80}} = -\frac{{1}}{{8}}$$

So, the solutions to the equation $27x + 4 = 40x^2$ using the quadratic formula are $x = \frac{{4}}{{5}}$ and $x = -\frac{{1}}{{8}}$.