How do you find the roots, real and imaginary, of #y=10x^2 + 9x-1 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y=10x^2 + 9x-1         # using the quadratic formula?

Isaiah Anthony · Accepted Answer

#x_(1,2)=(-B+-sqrt(B^2-4AC))/(2A)# #x_(1,2)=(-9+-sqrt((-9)^2-4*10*(-1)))/(2*10)# #x_(1,2)=(-9+-sqrt(81+40))/(20)# #x_(1,2)=-9/20+-1/20sqrt121=-9/20+-11/20# #->x_1=-9/20+11/20=2/20=1/10# #->x_2=-9/20-11/20=-20/20=-1#

Eli Assouline · Answer

undefined To find the roots of the quadratic equation $y = 10x^2 + 9x - 1$, we can use the quadratic formula:

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

where $a$, $b$, and $c$ are the coefficients of the quadratic equation in the form $ax^2 + bx + c = 0$.

In this case, $a = 10$, $b = 9$, and $c = -1$. Plugging these values into the formula, we get:

$$x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 10 \cdot (-1)}}{2 \cdot 10}$$

Simplifying under the square root gives:

$$x = \frac{-9 \pm \sqrt{81 + 40}}{20}$$

$$x = \frac{-9 \pm \sqrt{121}}{20}$$

$$x = \frac{-9 \pm 11}{20}$$

This gives us two roots:

$$x_1 = \frac{-9 + 11}{20} = \frac{2}{20} = \frac{1}{10}$$

$$x_2 = \frac{-9 - 11}{20} = \frac{-20}{20} = -1$$

So, the roots of the equation are $x = \frac{1}{10}$ and $x = -1$.