How do you solve $x^{2} + 21 x + 110 = 0$ $x^2+21x+110=0$ ?

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How do you solve $x^{2} + 21 x + 110 = 0$ $x^2+21x+110=0$ ?

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Answer 1 · 2016-03-23T13:03:41

$x = - 10$ $x= -10$ AND $x = - 11$ $x = -11$

Explanation:

$x^{2} + 21 x + 110 = 0$ $x^2 +21x+110= 0$ The trinomial in this equation can be factored into the product of two binomials. Look for two numbers which multiply to give a product of 110 and have a sum of 21. In this case the numbers are $10$ $10$ and $11$ $11$ .

$(10) (11) = 110$ $(10)(11) =110$ and $10 + 11 = 21$ $10+11=21$

So, $x^{2} + 21 x + 110 = 0$ $x^2+21x+110= 0$ can be rewritten as:

$(x + 10) (x + 11) = 0$ $(x+10)(x+11)=0$ . Now either

$(x + 10) = 0$ $(x+10) =0$ and therefore $x = - 10$ $x=-10$

or $(x + 11) = 0$ $(x+11)=0$ and $x = - 11$ $x=-11$

Answer 2 · 2016-03-23T14:12:31

$x = - 10, - 11$ $x=-10,-11$

Explanation:

$x^{2} + 21 x + 110 = 0$ $color(blue)(x^2+21x+110=0$

You can solve this both by factoring and Quadratic formula

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

Factoring

If you have a problem with factoring

Watch this video:

Factor the equation

$\to (x + 10) + (x + 12) = 0$ $rarr(x+10)+(x+12)=0$

Now we can say

$x + 10 = 0, x + 11 = 0$ $color(orange)(x+10=0,x+11=0$

$\Rightarrow x = - 10, - 11$ $color(green)(rArrx=-10,-11$

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

This is a Quadratic equation (in form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ )

Quadratic formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

Where

$a = 1, b = 21, c = 110$ $color(red)(a=1,b=21,c=110$

$\to x = \frac{- 21 \pm \sqrt{21^{2} - 4 (1) (110)}}{2 (1)}$ $rarrx=(-21+-sqrt(21^2-4(1)(110)))/(2(1))$

$\to x = \frac{- 21 \pm \sqrt{21^{2} - 4 (110)}}{2}$ $rarrx=(-21+-sqrt(21^2-4(110)))/(2)$

$\to x = \frac{- 21 \pm \sqrt{441 - 440}}{2}$ $rarrx=(-21+-sqrt(441-440))/(2)$

$\to x = \frac{- 21 \pm \sqrt{1}}{2}$ $rarrx=(-21+-sqrt(1))/(2)$

$\to x = \frac{- 21 \pm 1}{2}$ $rarrx=(-21+-1)/(2)$

Now we have two solutions

$1) \frac{- 21 + 1}{2} = - \frac{20}{2} = - 10$ $color(indigo)(1))color(indigo)((-21+1)/2=-20/2=-10$

$2) \frac{- 21 - 1}{2} = - \frac{22}{2} = - 11$ $color(orange)(2))color(orange)((-21-1)/2=-22/2=-11$

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

$∴ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ | x = - 10, - 11 | - ------------- -$ $color(blue)( :.ul bar |x=-10,-11|$

Answer 3 · 2016-03-23T15:33:55

(x + 10)(x + 11)

Explanation:

$y = x^{2} + 21 x + 110 = 0$ $y = x^2 + 21x + 110 = 0$
Use the new AC Method to factor trinomials (Socratic Search).
Find 2 numbers knowing sum (b = 21) and product (c = 11).
Since ac > 0, they have same sign.
Compose factor pairs of (c = 110) --> ...(5, 22)(10, 11). This sum is 21 = b. Then the numbers are 10 and 11.
Answer: y = (x + 10)(x + 11)

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