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How do you solve \[3{{x}^{2}}=-12x-15\] ?

Last updated date: 13th Jun 2024
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Hint: We will rearrange the equation and will solve the equation as a quadratic equation and will simplify the equation by finding the roots and by identifying which formula to use.

Complete step by step answer:
We have the given equation:
Now on transferring terms and rearranging we get:
Now for a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ , we will use the following formula: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Therefore from the above equation and on comparing, we get $a=3,b=12,c=15$
Now, substituting the values in formula we get: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-12\pm \sqrt{{{12}^{2}}-4*3*15}}{2*3}$
Now we will first simplify, root part in our equation:
$\Rightarrow \sqrt{144-180}$
$\Rightarrow \sqrt{-36}$
Now by applying radical rule, i.e. $\sqrt{-a}=\sqrt{-1}\sqrt{a}$ we get:
$\Rightarrow \sqrt{-1}\sqrt{36}$
Now we can see we can apply imaginary rule: $\sqrt{-1}=i$
An imaginary number is a number that is expressed in terms of square root of a negative number (usually the square root of $-1$, and is represented by $i$ or $j$ )
Therefore, we get:
$\Rightarrow \sqrt{36}i$
On simplifying, we get:
$\Rightarrow 6i$
Now coming to our equation and substituting what we equate, we get:
\[\] $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-12\pm 6i}{6}$
Now on factorization, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{6\left( -2\pm i \right)}{6}$
On dividing and equating, we get:
$\left( {{x}_{1}},{{x}_{2}} \right)=-2\pm i$
Now on equating the above equation we get the value of x, i.e.:
${{x}_{1}}=-2+i$ and ${{x}_{2}}=-2-i$
Therefore, the solution to the given quadratic equation are:
${{x}_{1}}=-2+i$ and ${{x}_{2}}=-2-i$

There are three forms of quadratic equation, Standard form:$y=a{{x}^{2}}+bx+c$ where $a,b,c$ are just numbers. Factored form:\[~y\text{ }=\text{ }\left( ax\text{ }+\text{ }c \right)\left( bx\text{ }+\text{ }d \right)\] again the $a,b,c,d$ are just numbers. Vertex form: \[y\text{ }=\text{ }a{{\left( x\text{ }+\text{ }b \right)}^{2}}\text{ }+\text{ }c\] again the $a,b,c$ are just numbers. We have to identify by looking at the equation which form we have to use, then only we can solve the equation. If we have the standard form of the quadratic equation, we can verify the answer by substituting the answer in the original equation.