How do you solve #x^2 – 10x – 1 = -10# using the quadratic formula?

Question

Eva Adamson · Accepted Answer

#x=9,1#

Given:

#x^2-10x-1=-10#

All terms should be moved to the left.

#x^2-10x-1+10=0#

Simplify.

#x^2-10x+9# #larr# Quadratic equation in standard form:

#ax^2+bx+c=0#, where:

#a=1#, #b=-10#, and #c=9#.

formula for quadratics

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

Enter the specified values in the formula.

#x=(-(-10)+-sqrt((-10)^2-4*1*9))/(2*1)#

Simplify.

#x=10+-sqrt(100-36))/2#

Simplify.

#x=(10+-sqrt64)/2#

#x=(10+-8)/2#

Solutions for #x#.

#x=(10+8)/2,# #(10-8)/2#

Simplify.

#x=18/2,# #2/2#

#x=9,1#

Genesis Allen · Answer

undefined To solve the equation $x^2 - 10x - 1 = -10$ using the quadratic formula:

1. First, rewrite the equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -10$, and $c = -1 + 10 = 9$.

2. Substitute the values of $a$, $b$, and $c$ into the quadratic formula: 
$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

3. Plug in the values:
$$x = \frac{{-(-10) \pm \sqrt{{(-10)^2 - 4(1)(9)}}}}{{2(1)}}$$

4. Simplify inside the square root:
$$x = \frac{{10 \pm \sqrt{{100 - 36}}}}{{2}}$$
$$x = \frac{{10 \pm \sqrt{{64}}}}{{2}}$$
$$x = \frac{{10 \pm 8}}{{2}}$$

5. Compute the two possible values for $x$:
$$x_1 = \frac{{10 + 8}}{{2}} = 9$$
$$x_2 = \frac{{10 - 8}}{{2}} = 1$$

Thus, the solutions to the equation are $x = 9$ and $x = 1$.