 Michael Maggard

2021-12-27

To solve:
${x}^{2}+11x+24=0$ Deufemiak7

Step 1
Given: ${x}^{2}+11x+24=0$
Split the middle term 11x as $+3x$ and $+8x$
Replace 11x with $+3x+8x$ in ${x}^{2}+11x+24=0$
${x}^{2}+8x+3x+24=0$
The above equation can be written ads $x×x+8×x+3×x+3×8$
Apply distributive property for first two terms and last two terms,
$x×x+8×x+3×x+3×8=x\left(x+8\right)+3\left(x+8\right)$
Again apply distributive property for $x\left(x+8\right)+3\left(x+8\right)$
$x\left(x+8\right)+3\left(x+8\right)=\left(x+3\right)\left(x+8\right)$
So the factors are $\left(x+8\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(x+3\right)$
${x}^{2}+11x+24=\left(x+8\right)\left(x+3\right)$
To find the value of x
Take $x+8=0$
Subtract 8 on both sides,
$x+8-8=0-8$ NSk $x=-8$
Take $x+3=0$
Subtract 3 on both sides,
$x-3+3=0-3$
$x=-3$
So, the values of x are -8 and -3 Vivian Soares

Step 1
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form
${x}^{2}+bx=c$
${x}^{2}+11x+24=0$
Subtract 24 from both sides of the equation.
${x}^{2}+11x+24-24=-24$
Subtracting 24 from itself leaves 0
${x}^{2}+11x=-24$
Divide 11, the coefficient of the x term, by 2 to get $\frac{11}{2}$. Then add the square of $\frac{11}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
${x}^{2}+11x+{\left(\frac{11}{2}\right)}^{2}=-24+{\left(\frac{11}{2}\right)}^{2}$
Square $\frac{11}{2}$ by squaring both the numerator and the denominator of the fraction.
${x}^{2}+11x+\frac{121}{4}=-24+\frac{121}{4}$
Add -24 to $\frac{121}{4}$
${x}^{2}+11x+\frac{121}{4}=\frac{25}{4}$
Factor ${x}^{2}+11x+\frac{121}{4}$. In general, when ${x}^{2}+bx+c$ is a perfect square, it can always be factored as ${\left(x+\frac{b}{2}\right)}^{2}$
${\left(x+\frac{11}{2}\right)}^{2}=\frac{25}{4}$
Take the square root of both sides of the equation.
$\sqrt{{\left(x+\frac{11}{2}\right)}^{2}}=\sqrt{\frac{25}{4}}$
Simplify.
$x+\frac{11}{2}=\frac{5}{2}$
$x+\frac{11}{2}=-\frac{5}{2}$
Subtract $\frac{11}{2}$ from both sides of the equation.
$x=-3$
$x=-8$ karton

Step 1
Determine the quadratic equation's coefficients a, b and c
Use the standard form, $a{x}^{2}+bx+c=0$, to find the coefficients of our equation, ${x}^{2}+11x+24=0$
a=1
b=11
c=24
Step 2
Plug these coefficients into the quadratic formula
The quadratic formula gives us the roots for $a{x}^{2}+bx+c=0$, in which a, b and c are numbers (or coefficients), as follows:a=1
b=11
c=24

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{-11±\sqrt{{11}^{2}-4×1×24}}{2×1}$
Simplify exponents and square roots
$\begin{array}{}x=\frac{-11±\sqrt{121-4×1×24}}{2×1}\\ \text{Perform any multiplication or division, from left to right:}\\ x=\frac{-11±\sqrt{121-4×24}}{2×1}\\ x=\frac{-11±\sqrt{121-96}}{2×1}\\ x=\frac{-11±\sqrt{25}}{2×1}\\ x=\frac{-11±\sqrt{25}}{2}\end{array}$
to get the result:
$x=\frac{-11±\sqrt{25}}{2}$
Step 3
Simplify square root $\sqrt{25}$
Simplify 25 by finding its prime factors
The prime factorization of 25 is ${5}^{2}$
Write the prime factors:
$\sqrt{25}=\sqrt{5×5}$
Group the prime factors into pairs and rewrite them in exponent form:
$\sqrt{5×5}=\sqrt{{5}^{2}}$
Use the rule $\sqrt{{x}^{2}}=x$ to simplify further:
$\sqrt{{5}^{2}}=5$
Step 4
Solve the equation for x
$x=\frac{-11±5}{2}$
The $±$ means two answers are possible
Separate the equations: ${x}_{1}=\frac{-11+5}{2}$ and ${x}_{2}=\frac{-11-5}{2}$
$\begin{array}{}{x}_{1}=\frac{-11+5}{2}\\ {x}_{1}=\frac{-6}{2}\\ {x}_{1}=-3\\ {x}_{2}=\frac{-11-5}{2}\\ {x}_{2}=\frac{-16}{2}\\ {x}_{2}=-8\end{array}$

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