Question

Solve:$25{{x}^{2}}-30x+9=0$

Answer 1

Hint: To solve the quadratic equation means we need to find the roots of the quadratic equation. So we can find out the quadratic equation by using the Shri Dharacharya formula, which states as$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step answer:
Moving ahead with the question in step wise manner;
As quadratic equation is given as$25{{x}^{2}}-30x+9=0$
So as we know that to find out the roots of any quadratic equation we can find it through the formula called Shri Dharacharya. Which says that when a quadratic equation$a{{x}^{2}}+bx+c=0$exist then the roots of this quadratic equation will be$\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$and$\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$which can also be written as$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
So to find the roots of the quadratic equation given in our question$25{{x}^{2}}-30x+9=0$, let us first compare it with general form of equation$a{{x}^{2}}+bx+c=0$. So by comparing we can say that;
$\begin{align}
  & a=25 \\
 & b=-30 \\
 & c=9 \\
\end{align}$
So to find the roots of quadratic equation put the value of a, b, c in the Shri Dharacharya formula; so it will give us;
$\begin{align}
  & \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \dfrac{-\left( -30 \right)\pm \sqrt{{{\left( -30 \right)}^{2}}-4\left( 25 \right)\left( 9 \right)}}{2\left( 25 \right)} \\
\end{align}$
Upon simplifying it we will get;
\[\dfrac{30\pm \sqrt{{{\left( -30 \right)}^{2}}-900}}{50}\]
As we know that square of$-30$is 900 i.e.${{\left( -30 \right)}^{2}}=900$, so we can write above equation as;
\[\dfrac{30\pm \sqrt{900-900}}{50}\]
On further solving we will get;
\[\dfrac{30\pm \sqrt{0}}{50}\]
And as we also know that$\sqrt{0}=0$so we can write above equation as;
\[\dfrac{30\pm 0}{50}\],
As we know that we can write$a\pm b$as$a+b$and$a-b$, so in the above equation we can write it in two different numbers i.e.
\[\dfrac{30+0}{50}\]and\[\dfrac{30-0}{50}\]
On solving both equation separately they will give us;
\[\dfrac{30}{50}=\dfrac{3}{5}\]and\[\dfrac{30}{50}=\dfrac{3}{5}\],
As both roots are the same, we can say that the roots of quadratic equations are the same which are equal to\[\dfrac{3}{5}\] .

Hence the answer is\[\dfrac{3}{5}\].

Note: We can find the roots of quadratic equation by splitting middle term method, which is more simple to use when the value of ‘a’ is 1 in equation$a{{x}^{2}}+bx+c=0$, or when we have to find factors, otherwise if we want to find only roots in quadratic equation then Shri Dharacharya formula is best option.