Question

How do you solve $3x\left( -x+3 \right)=15$ ?

Answer 1

Hint: We can easily solve problems on quadratic equations like this by completing a square method. We will simplify the given equation to get a quadratic equation in the simplest form and rearrange some terms to have a complete square in the form of ${{\left( x-a \right)}^{2}}$ . Now, we further simplify the equation to find the solution of the equation.

Complete step by step answer:
The given equation is
$3x\left( -x+3 \right)=15$
We use the distributive property in the terms of the left-hand side of the above equation we get
$\Rightarrow -3{{x}^{2}}+9x=15$
We multiply both the sides of the above equation with $\left( -1 \right)$ and get
$\Rightarrow 3{{x}^{2}}-9x=-15$
Now, dividing both the sides of the above equation by $3$ and get
$\Rightarrow {{x}^{2}}-3x=-5$
We rewrite the above equation as
$\Rightarrow {{x}^{2}}-2\cdot \dfrac{3}{2}x=-5....\text{expression}1$
To have a complete square in the left-hand side let’s take the square ${{\left( x-a \right)}^{2}}$ for comparison.
We know that, ${{\left( x-a \right)}^{2}}={{x}^{2}}-2\cdot a\cdot x+{{a}^{2}}....\text{expression2}$
As, the first two terms in the expression $\left( {{x}^{2}}-2\cdot a\cdot x+{{a}^{2}} \right)$ are ${{x}^{2}}$ and $2x$ , we compare the left-hand side of $\text{expression}1$ with the right-hand side of $\text{expression2}$ and get the value of $a$ as $a=\dfrac{3}{2}$ .
Hence, to get the square term ${{\left( x-\dfrac{3}{2} \right)}^{2}}$ we add $\dfrac{9}{4}$ to the both sides of $\text{expression}1$
$\Rightarrow {{x}^{2}}-2\cdot \dfrac{3}{2}x+\dfrac{9}{4}=-5+\dfrac{9}{4}$
$\Rightarrow {{x}^{2}}-2\cdot \dfrac{3}{2}x+\dfrac{9}{4}=-\dfrac{11}{4}$
The above equation can be also written as
$\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{-11}{4}$
Now, taking square root on both the sides of the equation, we get
$\Rightarrow x-\dfrac{3}{2}=\pm \dfrac{\sqrt{\left( -11 \right)}}{2}$
The square root of $\left( -1 \right)$ is taken as imaginary unit $i$ as shown below
$\Rightarrow x-\dfrac{3}{2}=\pm \dfrac{i\sqrt{11}}{2}$
Further simplifying the above equation, we get
$\Rightarrow x=\dfrac{i\sqrt{11}}{2}+\dfrac{3}{2}$
$\Rightarrow x=\dfrac{i\sqrt{11}+3}{2}$
And $\Rightarrow x=-\dfrac{i\sqrt{11}}{2}+\dfrac{3}{2}$
$\Rightarrow x=-\dfrac{i\sqrt{11}-3}{2}$
Therefore, the solution of the equation $3x\left( -x+3 \right)=15$ are $x=\dfrac{i\sqrt{11}+3}{2}$ and $x=-\dfrac{i\sqrt{11}-3}{2}$

Note:
While square rooting the negative terms, we must keep in mind that the square root of $\left( -1 \right)$ is taken as an imaginary unit $i$ so that accurate solutions are obtained. Also, the given equation can be solved using other methods such as applying the Sridhar Acharya formula, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $a=1,\text{ }b=-3\text{ and }c=5$ .