Question

How do you solve $5{{x}^{2}}=125$? \[\]

Answer 1

Hint: We see that the given equation is a quadratic equation . We convert the given equation with the general quadratic equation $a{{x}^{2}}+bx+c=0$and use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the solutions. We alternatively use factorization polynomials to find the solutions. \[\]

Complete step by step answer:
We know that the quadratic equation in one variable $x$ is given by $a{{x}^{2}}+bx+c=0$ where $a\ne 0,b,c$ are real numbers. The real solution otherwise called roots for the quadratic equation exists when the discriminant $D={{b}^{2}}-4ac\ge 0$. We also know that the roots of the equation are given by the formula \[\]
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
The given quadratic equation is
\[5{{x}^{2}}=125\]
 We can write in the general form as
\[5{{x}^{2}}-125=0\]
Let us check whether the given quadratic equation has real roots or not. We compare the given equation with the general quadratic equation $a{{x}^{2}}+bx+c=0$ and find $a=5,b=0,c=-125$. So the value of the discriminant is $D={{b}^{2}}-4ac={{0}^{2}}-4\left( 5 \right)\left( -125 \right)=4\times 625>0$ . So there are real roots for the given quadratic equation. We use quadratic formula and find the roots as
\[x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\cdot 5\cdot \left( -125 \right)}}{2\times 5}=\dfrac{\pm \sqrt{4\cdot 5\cdot 125}}{2\times 5}=\dfrac{\pm \sqrt{4\cdot 5\cdot 5\cdot 5\cdot 5}}{2\times 5}=\dfrac{\pm 2\times 5\times 5}{2\times 5}=\pm 5\]
 So we get two roots of the equation which are additive inverse of each other as $x=5,-5$.
Alternative method: We can alternatively solve using factorization. Let us consider
\[\begin{align}
  & 5{{x}^{2}}=125 \\
 & \Rightarrow 5{{x}^{2}}-125=0 \\
\end{align}\]
We take 5 common from both the term in the left hand side to have
\[\begin{align}
  & \Rightarrow 5\left( {{x}^{2}}-25 \right)=0 \\
 & \Rightarrow 5\left( {{x}^{2}}-{{5}^{2}} \right)=0 \\
\end{align}\]
We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=x,b=5$ in the above step to have ;
\[\Rightarrow 5\left( x+5 \right)\left( x-5 \right)=0\]
We know that if product of three factors is zero then at least one of them is zero .Since $5\ne 0$then we have
\[\begin{align}
  & \Rightarrow x+5=0\text{ or }x-5=0 \\
 & \Rightarrow x=-5\text{ or }x=5 \\
\end{align}\]

So the solution of the given equation is same $x=5,-5$.\[\]

Note: We note that if discriminant $D=0$ , we get two equal roots. If D is a perfect square we get two rational roots otherwise two conjugate irrational roots. If D is a perfect square and $b=0$ we get two roots which are additive inverse of each other. If $-b\pm \sqrt{D}$ is a multiple of $2a$ we get integral roots.