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How do you solve the quadratic using the quadratic formula given $3{{a}^{2}}=6a-3$?

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Last updated date: 21st Jun 2024
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Answer
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Hint: We have been given a quadratic equation of $a$ as $3{{a}^{2}}=6a-3$. We use the quadratic formula to solve the value of the $a$. we have the solution in the form of $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$ for general equation of $p{{x}^{2}}+qx+r=0$. We put the values and find the solution.

Complete step-by-step solution:
We know for a general equation of quadratic $p{{x}^{2}}+qx+r=0$, the value of the roots of $x$ will be $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$. This is the quadratic equation solving method. The root part $\sqrt{{{q}^{2}}-4pr}$ of $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$ is called the discriminant of the equation.
In the given equation we have $3{{a}^{2}}-6a+3=0$. The values of p, q, r are $3,-6,3$ respectively.
We put the values and get $a$ as \[a=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 3\times 3}}{2\times 3}=\dfrac{6\pm 0}{6}=\dfrac{6}{6}=1\]
The roots of the equation are real numbers. Two roots being equal in value as the expression is a square form. So, values of $a$ are $a=1$.

Note: We can also apply the middle-term factoring or grouping to factorise the polynomial.
We have
$\begin{align}
  & 3{{a}^{2}}=6a-3 \\
 & \Rightarrow {{a}^{2}}-2a+1=0 \\
 & \Rightarrow {{\left( a-1 \right)}^{2}}=0 \\
\end{align}$
Now we take square root both sides of the equation and get
$\begin{align}
  & {{\left( a-1 \right)}^{2}}=0 \\
 & \Rightarrow a-1=0 \\
\end{align}$
The root of the equation becomes 1 where two roots being equal in value.
Therefore, values of $a$ are $a=1$.