
How do you solve $9{x^2} + 11x + 18 = - 10x + 8$?
Answer
548.4k+ views
Hint:The given equation is a quadratic equation in one variable $x$. The general form of a quadratic equation is given by $a{x^2} + bx + c = 0$. Solving this equation gives two values of the variable $x$ as the result. We will use the factorization method to solve this quadratic equation.
Complete solution step by step:
We have to solve the given equation $9{x^2} + 11x + 18 = - 10x + 8$. To find the value of $x$, we have to put the values of $a$, $b$ and $c$ in the quadratic formula. To get the values of $a$, $b$ and $c$ from the given equation, we rearrange the equation and compare it with the general form of the quadratic equation.
$
9{x^2} + 11x + 18 = - 10x + 8 \\
\Rightarrow 9{x^2} + 11x + 10x + 18 - 8 = 0 \\
\Rightarrow 9{x^2} + 21x + 10 = 0 \\
$
First step of the factorization method involves separation of the expression in the form $bx$ as $px$ and $qx$ such that $(p + q) = b$ and $ac = pq$.
Then by hit and trial method, we find that $p = 15$ and $q = 6$ would satisfy $(p + q) = b$ and $ac = pq$.
Thus, we get:
$ \Rightarrow 9{x^2} + 15x + 6x + 10 = 0$
Then, we take $3x$ common from the first two terms and $2$ from the last two terms on the Left Hand
Side or LHS.
$ \Rightarrow 3x(3x + 5) + 2(3x + 5) = 0$
Now we take the expression $(3x + 5)$common on the LHS.
$ \Rightarrow (3x + 5)(3x + 2) = 0$
From the above equation, we get:
$
\Rightarrow 3x + 5 = 0 \\
\Rightarrow 3x = - 5 \\
\Rightarrow x = \dfrac{{ - 5}}{3} \\
$
Also, we have:
$
3x + 2 = 0 \\
3x = - 2 \\
x = \dfrac{{ - 2}}{3} \\
$
Hence, the two required values of $x$ that we get on solving the given equation are $\dfrac{{ - 5}}{3}$ and $\dfrac{{ - 2}}{3}$.
Note: Another method to solve for $x$ in the quadratic equation is by using the quadratic formula. The quadratic formula to solve for such equation is given by, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where $a$, $b$ and $c$ are the numerical values of coefficients in the quadratic equation. We can check the answer by putting the result in the given equation to satisfy LHS = RHS.
Complete solution step by step:
We have to solve the given equation $9{x^2} + 11x + 18 = - 10x + 8$. To find the value of $x$, we have to put the values of $a$, $b$ and $c$ in the quadratic formula. To get the values of $a$, $b$ and $c$ from the given equation, we rearrange the equation and compare it with the general form of the quadratic equation.
$
9{x^2} + 11x + 18 = - 10x + 8 \\
\Rightarrow 9{x^2} + 11x + 10x + 18 - 8 = 0 \\
\Rightarrow 9{x^2} + 21x + 10 = 0 \\
$
First step of the factorization method involves separation of the expression in the form $bx$ as $px$ and $qx$ such that $(p + q) = b$ and $ac = pq$.
Then by hit and trial method, we find that $p = 15$ and $q = 6$ would satisfy $(p + q) = b$ and $ac = pq$.
Thus, we get:
$ \Rightarrow 9{x^2} + 15x + 6x + 10 = 0$
Then, we take $3x$ common from the first two terms and $2$ from the last two terms on the Left Hand
Side or LHS.
$ \Rightarrow 3x(3x + 5) + 2(3x + 5) = 0$
Now we take the expression $(3x + 5)$common on the LHS.
$ \Rightarrow (3x + 5)(3x + 2) = 0$
From the above equation, we get:
$
\Rightarrow 3x + 5 = 0 \\
\Rightarrow 3x = - 5 \\
\Rightarrow x = \dfrac{{ - 5}}{3} \\
$
Also, we have:
$
3x + 2 = 0 \\
3x = - 2 \\
x = \dfrac{{ - 2}}{3} \\
$
Hence, the two required values of $x$ that we get on solving the given equation are $\dfrac{{ - 5}}{3}$ and $\dfrac{{ - 2}}{3}$.
Note: Another method to solve for $x$ in the quadratic equation is by using the quadratic formula. The quadratic formula to solve for such equation is given by, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where $a$, $b$ and $c$ are the numerical values of coefficients in the quadratic equation. We can check the answer by putting the result in the given equation to satisfy LHS = RHS.
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