Question

How do you solve \[30{{x}^{2}}+5x-10=0\]?

Answer 1

Hint: This question is from the topic of quadratic equation. In this question, we have to find the value of x. For solving this question, we will first divide the whole equation by 5. After that, we will solve the quadratic equation using the solving method of Sridharacharya’s rule and will get the value of x. After that, we will see the alternate method to solve this question.

Complete step by step answer:
Let us solve this question.
In this question, we have to solve the given term \[30{{x}^{2}}+5x-10=0\]. That means, we have to find the value of x from the given equation \[30{{x}^{2}}+5x-10=0\].
The equation we have to solve is
\[30{{x}^{2}}+5x-10=0\]
Dividing both sides of equation by 5, we get
\[\Rightarrow \dfrac{30{{x}^{2}}+5x-10}{5}=\dfrac{0}{5}\]
The above equation can also be written as
\[\Rightarrow \dfrac{30{{x}^{2}}}{5}+\dfrac{5x}{5}-\dfrac{10}{5}=\dfrac{0}{5}\]
As we know that 30 divided by 5 can be written as 6, 5 divided by 5 can be written as 1, 10 divided by 5 can be written as 2 and 0 divided by 5 can be written as 0, so we can write the above equation as
\[\Rightarrow 6{{x}^{2}}+x-2=0\]
Now, we will find the value of x from the above equation using Sridharacharya’s rule.
According to Sridharacharya’s rule, we can write the value of x as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 6\times \left( -2 \right)}}{2\times 6}\]
The above can also be written as
\[x=\dfrac{-1\pm \sqrt{1+48}}{12}=\dfrac{-1\pm \sqrt{49}}{12}=\dfrac{-1\pm 7}{12}\]
So, we can say that the values of x will be \[\dfrac{-1+7}{12}\] and \[\dfrac{-1-7}{12}\]
Or, we can say that the values of x are \[\dfrac{6}{12}\] and \[\dfrac{-8}{12}\]
Or, we can say that the values of x are \[\dfrac{1}{2}\] and \[\dfrac{-2}{3}\].
Hence, we have solved the equation \[30{{x}^{2}}+5x-10=0\], and found the values of x as \[\dfrac{1}{2}\] and \[\dfrac{-2}{3}\].

Note:
We should have a better knowledge in the topic of algebra and quadratic equations to solve this type of question easily. We should know about Sridharacharya’s rule. This rule says that if we have given an equation as \[a{{x}^{2}}+bx+c=0\], then the value of x from this equation will be
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Let us solve this question by a different method.
The equation we have to solve is \[30{{x}^{2}}+5x-10=0\] can also be written as
\[\Rightarrow 6{{x}^{2}}+x-2=0\]
Separate the coefficient of term x so that if we write them and multiply then it comes equal to multiplication of 6 and (-2).
We can write the equation as
\[\Rightarrow 6{{x}^{2}}+\left( 4-3 \right)x-2=0\]
The above equation can also be written as
\[\Rightarrow 6{{x}^{2}}+4x-3x-2=0\]
Now, we can write the above equation as
\[\Rightarrow 2x\left( 3x+2 \right)-1\left( 3x+2 \right)=0\]
The above equation can also be written as
\[\Rightarrow \left( 2x-1 \right)\left( 3x+2 \right)=0\]
We can write the above as
\[2x-1=0\] and \[3x+2=0\]
From the above, we can say that
\[x=\dfrac{1}{2}\] and \[x=-\dfrac{2}{3}\]
So, we can use this method too to solve this question.