How do you find the zeros, real and imaginary, of #y=3x^2+17x-3# using the quadratic formula?

Question

Eva Abramson · Accepted Answer

The quadratic has 2 real roots at #x_+=(5sqrt(13)-17)/6~=0.171# and #x_(-)=(-17-5sqrt(13))/6~=-5.84#

To begin, we must put our quadratic in the standard form (which is it already) and set #y# equal to zero.

#3x^2+17x-3=0#

The quadratic formula uses the form:

#ax^2 +bx+c=0#

where a,b, and c can be gotten from the equation above. The roots are given by:

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

If the roots are complex, we will get a negative number under the square root. In our case the roots are given by:

#x=(-17+- sqrt(17^2 -4*3*(-3)))/ (2*3)#

#x = (-17+- 5sqrt(13))/ (6)#

#x_+=(5sqrt(13)-17)/6~=0.171#

#x_(-)=(-17-5sqrt(13))/6~=-5.84#

Eva Abramson · Answer

undefined To find the zeros of $y = 3x^2 + 17x - 3$ using the quadratic formula ($x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$):

1. Identify the values of $a$, $b$, and $c$:
   - $a = 3$
   - $b = 17$
   - $c = -3$

2. Substitute these values into the quadratic formula:
   $$
   x = \frac{-17 \pm \sqrt{(17)^2 - 4(3)(-3)}}{2(3)}
   $$

3. Simplify the expression inside the square root:
   $$
   x = \frac{-17 \pm \sqrt{289 + 36}}{6}
   $$
   $$
   x = \frac{-17 \pm \sqrt{325}}{6}
   $$

4. Now, calculate the square root of 325:
   $$
   \sqrt{325} \approx 18.03
   $$

5. Substitute back into the formula:
   $$
   x = \frac{-17 \pm 18.03}{6}
   $$

6. Calculate the two possible solutions:
   - $x_1 = \frac{-17 + 18.03}{6} \approx \frac{1.03}{6} \approx 0.172$
   - $x_2 = \frac{-17 - 18.03}{6} \approx \frac{-35.03}{6} \approx -5.84$

So, the zeros of $y = 3x^2 + 17x - 3$ are $x \approx 0.172$ and $x \approx -5.84$.