Question

Find ‘a’ so that roots of \[{{x}^{2}}+2\left( 3a+5 \right)x+2\left( 9{{a}^{2}}+25 \right)=0\] are real.

Answer 1

Hint: As we know that a quadratic equation has real roots if its determinant (D) is greater than or equal to zero. The value of determinant (D) of a quadratic equation \[a{{x}^{2}}+bx+c=0\] is given by as follows:
\[D={{b}^{2}}-4ac\]. Using this, we can find the value of a.

Complete step-by-step answer:
We have been given the equation:
\[{{x}^{2}}+2\left( 3a+5 \right)x+2\left( 9{{a}^{2}}+25 \right)=0\]
Now on comparing \[{{x}^{2}}+2\left( 3a+5 \right)x+2\left( 9{{a}^{2}}+25 \right)=0\] with the equation \[a{{x}^{2}}+bx+c=0\], we will get as follows:
a = 1, b = \[2\left( 3a+5 \right)\] and c = \[2\left( 9{{a}^{2}}+25 \right)\]
Also, we know that if a quadratic equation has real roots, then its value of determinant (D) is greater than equal to zero.
The determinant(D) of the equation \[a{{x}^{2}}+bx+c=0\] is given as follows:
\[D={{b}^{2}}+4ac\]
Now we have a = 1, b = \[2\left( 3a+5 \right)\] and c = \[2\left( 9{{a}^{2}}+25 \right)\]
\[\begin{align}
  & D\ge 0 \\
 & {{b}^{2}}-4ac\ge 0 \\
 & {{\left[ 2\left( 3a+5 \right) \right]}^{2}}-4.1.2\left( 9{{a}^{2}}+25 \right)\ge 0 \\
 & 4\left( 9{{a}^{2}}+30a+25 \right)-8\left( 9{{a}^{2}}+25 \right)\ge 0 \\
\end{align}\]
Taking ‘4’ as common from the above inequality, we will get as follows:
\[4\left[ 9{{a}^{2}}+30a+25-2\left( 9{{a}^{2}}+25 \right) \right]\ge 0\]
On dividing both the sides of the equality of the equation by 4, we get as follows:
\[\begin{align}
  & \dfrac{4}{4}\left[ 9{{a}^{2}}+30a+25-18{{a}^{2}}-50 \right]\ge \dfrac{0}{4} \\
 & 9{{a}^{2}}+30a+25-18{{a}^{2}}-50\ge 0 \\
 & -9{{a}^{2}}+30a-25\ge 0 \\
\end{align}\]
Now taking all of the terms on the left hand side of the equation to the right hand side of the equation, we will get as follows:
\[\begin{align}
  & 0\ge 9{{a}^{2}}-30a+25 \\
 & 9{{a}^{2}}-30a+25\le 0 \\
 & {{\left( 3a \right)}^{2}}-2.5.\left( 3a \right)+{{\left( 5 \right)}^{2}}\le 0 \\
 & {{\left( 3a-5 \right)}^{2}}\le 0 \\
\end{align}\]
As we know that the square of any expression can’t be negative. So the inequality on the left hand side is equal to the right hand side only.
\[{{\left( 3a-5 \right)}^{2}}=0\]
On taking square root on both the sides of the equality of the equation, we get as follows:
\[\begin{align}
  & {{\left[ {{\left( 3a-5 \right)}^{2}} \right]}^{{}^{1}/{}_{2}}}=0 \\
 & 3a-5=0 \\
 & 3a=5 \\
 & a=\dfrac{5}{3} \\
\end{align}\]
Therefore, the value of ‘a’ is equal to \[\dfrac{5}{3}\] so that the given equation has real roots.

Note: Be careful while solving the inequality as there is a change of mistake. Also, remember the point if we multiply the inequality by a positive number, the sign of equality does not change. But in case of a negative number, then the sign of inequality gets reversed.