How do you solve #x^2+10x-9=0#?

Question

Eli Assouline · Accepted Answer

#x=-5+-sqrt(34)#

#x^2+10x-9=0#

is presented as:

#ax^2+bx+c=0#

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

To determine: Use the quadratic formula.

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

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

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

#color(white)(x) = (-10+-sqrt(136))/2#

#color(white)(x) = (-10+-sqrt(2^2*34))/2#

#color(white)(x) = (-10+-2sqrt(34))/2#

#color(white)(x) = -5+-sqrt(34)#

Jameson Allen · Answer

undefined To solve the equation $ x^2 + 10x - 9 = 0 $, you can use the quadratic formula:

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

In this equation, $ a = 1 $, $ b = 10 $, and $ c = -9 $. Plug these values into the formula and solve for $ x $.

Jameson Allen · Answer

undefined You can solve the quadratic equation $x^2 + 10x - 9 = 0$ 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 $ax^2 + bx + c = 0$.

For the equation $x^2 + 10x - 9 = 0$, the coefficients are $a = 1$, $b = 10$, and $c = -9$. Substituting these values into the quadratic formula:

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

$$x = \frac{{-10 \pm \sqrt{{100 + 36}}}}{{2}}$$

$$x = \frac{{-10 \pm \sqrt{{136}}}}{{2}}$$

$$x = \frac{{-10 \pm 2\sqrt{{34}}}}{{2}}$$

$$x = \frac{{-5 \pm \sqrt{{34}}}}{{1}}$$

Therefore, the solutions to the equation $x^2 + 10x - 9 = 0$ are $x = -5 + \sqrt{34}$ and $x = -5 - \sqrt{34}$.