How do you solve ${{x}^{2}}+1.5x=0$?
Answer
609.3k+ views
Hint: Since, the given equation is a simplest quadratic equation so, we will take out x as a common variable. And to solve it further we will keep the terms equal to 0 to get the right answer. After this, by the help of shifting process we will shift the constants to the right side of the equal sign along with correct positive or negative signs.
Complete step by step solution:
Before solving this question we will first get to know about the term quadratic equation. It is a type of equation which has the highest degree equals to 2. Moreover, the coefficient of the highest degree must not be equal to 1. For example, $a{{x}^{2}}+bx+c=0$ is a quadratic equation with the coefficient of highest degree not equal to 1.
Now, we know that the given equation ${{x}^{2}}+1.5x=0$ is a quadratic equation.
To solve this equation we are going to take x as a common variable and write it like so, $x\left( x+1.5 \right)=0$.
This clearly gives us either x = 0 or x + 1.5 = 0.
In the equation x + 1.5 = 0, we can keep the 1.5 number to the right side of the equation.
Thus, we will get either x = 0 or x = - 1.5.
Hence, the value of x is either 0 or – 1.5 only.
Note: There is also an alternative method to solve this equation. For this we will apply the formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to ${{x}^{2}}+1.5x=0$. Here, we will take a = 1, b = 1.5 and c = 0. Thus, we get,
\[\begin{align}
& x=\dfrac{-\left( 1.5 \right)\pm \sqrt{{{\left( 1.5 \right)}^{2}}-4\left( 1 \right)\left( 0 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{-1.5\pm \sqrt{{{\left( 1.5 \right)}^{2}}}}{2} \\
& \Rightarrow x=\dfrac{-1.5\pm 1.5}{2} \\
& \Rightarrow x=\dfrac{-1.5-1.5}{2},x=\dfrac{-1.5+1.5}{2} \\
& \Rightarrow x=\dfrac{-2\left( 1.5 \right)}{2},x=0 \\
& \Rightarrow x=-1.5,x=0 \\
\end{align}\]
Also, if we consider x + 1.5 = 0 then, instead of using shifting we could have subtracted 1.5 on both sides of the equation as x + 1.5 – 1.5 = 0 – 1.5. A lot of concentration is needed to solve these types of questions because we might get a wrong sigh as instead of getting – 1.5 we can get x = 1.5 which will be a wrong answer.
Complete step by step solution:
Before solving this question we will first get to know about the term quadratic equation. It is a type of equation which has the highest degree equals to 2. Moreover, the coefficient of the highest degree must not be equal to 1. For example, $a{{x}^{2}}+bx+c=0$ is a quadratic equation with the coefficient of highest degree not equal to 1.
Now, we know that the given equation ${{x}^{2}}+1.5x=0$ is a quadratic equation.
To solve this equation we are going to take x as a common variable and write it like so, $x\left( x+1.5 \right)=0$.
This clearly gives us either x = 0 or x + 1.5 = 0.
In the equation x + 1.5 = 0, we can keep the 1.5 number to the right side of the equation.
Thus, we will get either x = 0 or x = - 1.5.
Hence, the value of x is either 0 or – 1.5 only.
Note: There is also an alternative method to solve this equation. For this we will apply the formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to ${{x}^{2}}+1.5x=0$. Here, we will take a = 1, b = 1.5 and c = 0. Thus, we get,
\[\begin{align}
& x=\dfrac{-\left( 1.5 \right)\pm \sqrt{{{\left( 1.5 \right)}^{2}}-4\left( 1 \right)\left( 0 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{-1.5\pm \sqrt{{{\left( 1.5 \right)}^{2}}}}{2} \\
& \Rightarrow x=\dfrac{-1.5\pm 1.5}{2} \\
& \Rightarrow x=\dfrac{-1.5-1.5}{2},x=\dfrac{-1.5+1.5}{2} \\
& \Rightarrow x=\dfrac{-2\left( 1.5 \right)}{2},x=0 \\
& \Rightarrow x=-1.5,x=0 \\
\end{align}\]
Also, if we consider x + 1.5 = 0 then, instead of using shifting we could have subtracted 1.5 on both sides of the equation as x + 1.5 – 1.5 = 0 – 1.5. A lot of concentration is needed to solve these types of questions because we might get a wrong sigh as instead of getting – 1.5 we can get x = 1.5 which will be a wrong answer.
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