How do you use the quadratic formula to solve #9x^2-6x-35=0#?

Question

Madelyn Anderson · Accepted Answer

#x = 7/3" "# or #" "x = -5/3#

Given:

#9x^2-6x-35 = 0#

Be aware that this is written in standard form:

#ax^2+bx+c = 0#

with #a=9#, #b=-6# and #c=-35#

The quadratic formula yields the roots:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-(color(blue)(-6))+-sqrt((color(blue)(-6))^2-4(color(blue)(9))(color(blue)(-35))))/(2(color(blue)(9)))#

#color(white)(x) = (6+-sqrt(36+1260))/18#

#color(white)(x) = (6+-sqrt(1296))/18#

#color(white)(x) = (6+-36)/18#

#color(white)(x) = (1+-6)/3#

That is:

#x = 7/3" "# or #" "x = -5/3#

#color(white)()# Notes

Although the quadratic formula is very helpful and worthy of memorization, if you haven't already, I highly advise you to learn how to derive it from scratch.

Using the identity of the difference of squares, here's one method:

#A^2-B^2=(A-B)(A+B)#

Given:

#ax^2+bx+c = 0#

We are able to write:

#0 = ax^2+bx+c#

#color(white)(0) = a(x+b/(2a))^2+(c-b^2/(4a))#

#color(white)(0) = a((x+b/(2a))^2-(b^2-4ac)/(2a)^2)#

#color(white)(0) = a((x+b/(2a))^2-(sqrt(b^2-4ac)/(2a))^2)#

#color(white)(0) = a((x+b/(2a))-sqrt(b^2-4ac)/(2a))((x+b/(2a))+sqrt(b^2-4ac)/(2a))#

#color(white)(0) = a(x-(-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))#

Hence:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

Sadie Ackerman · Answer

undefined To use the quadratic formula to solve the equation $9x^2 - 6x - 35 = 0$, we first identify the coefficients $a$, $b$, and $c$ in the general quadratic equation $ax^2 + bx + c = 0$. Then, we substitute these values into the quadratic formula:

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

For the equation $9x^2 - 6x - 35 = 0$, $a = 9$, $b = -6$, and $c = -35$. Substituting these values into the quadratic formula, we get:

$$x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(9)(-35)}}}}{{2(9)}}$$

$$x = \frac{{6 \pm \sqrt{{36 + 1260}}}}{{18}}$$

$$x = \frac{{6 \pm \sqrt{{1296}}}}{{18}}$$

$$x = \frac{{6 \pm 36}}{{18}}$$

Now, we have two possible solutions:

$$x_1 = \frac{{6 + 36}}{{18}} = \frac{{42}}{{18}} = \frac{{7}}{{3}}$$

$$x_2 = \frac{{6 - 36}}{{18}} = \frac{{-30}}{{18}} = -\frac{{5}}{{3}}$$

So, the solutions to the equation $9x^2 - 6x - 35 = 0$ are $x = \frac{{7}}{{3}}$ and $x = -\frac{{5}}{{3}}$.