Question

Solve the quadratic equation ${{x}^{2}}-196=0$$$$$

Answer 1

We compare 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 roots. We alternatively solve by factorizing the quadratic polynomial ${{x}^{2}}-196$ in the given equation using factor theorem. We first find one root with trial and error as $x=14$ and then divide ${{x}^{2}}-196$ by $x-14$ to find the other factor.

Complete step-by-step solution:
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 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
${{x}^{2}}-196=0$
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 $a=1,b=0,c=-196$. So the value of the discriminant is $D={{b}^{2}}-4ac={{0}^{2}}-4\left( 1 \right)\left( -196 \right)=784>0$ . So real roots for the given quadratic equation exist. The roots are

& x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 1\times \left( -196 \right)}}{2\times 1} \\\ & =\dfrac{\pm \sqrt{4\times 196}}{2} \\\ & =\dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 7}}{2} \\\ & =\dfrac{\pm \left( 2\times 2\times 7 \right)}{2} \\\ & =\pm 14 \\\ \end{aligned}$$ We can alternatively solve the problem by factorizing the quadratic polynomial the given quadratic equation . We know from the factor theorem that a polynomial $p\left( x \right)$ has a factor $\left( x-a \right)$ if and only if $p\left( a \right)=0$ in other words $a$ is a zero of $p\left( x \right)$ . So we first find a root by trial and error. Let us put $x=14$ in the quadratic polynomial say $p\left( x \right)={{x}^{2}}-196$ . We get $p\left( 14 \right)=0$. So $x-14$ is a factor of $p\left( x \right)$ . Now we need to find the other root by dividing $p\left( x \right)$ by $x-14$ using long division method. $$\begin{matrix} x-14 & \overset{x+14}{\overline{\left){{{x}^{2}}-196}\right.}} \\\ {} & \underline{-\left( {{x}^{2}}-14x \right)} \\\ {} & 14x-196 \\\ {} & \underline{-\left( 14x-196 \right)} \\\ {} & 0 \\\ \end{matrix}$$ So we have the other factor at the quotient of the division as $x+14$. We proceed by replacing $p\left( x \right)={{x}^{2}}-196$ with $\left( x-14 \right)\left( x+14 \right)$ and get $$\begin{aligned} & p\left( x \right)=\left( x-14 \right)\left( x+14 \right)=0 \\\ & \Rightarrow x=14,-14 \\\ \end{aligned}$$ So we got two real roots $x=14$ and $x=-14$ whose absolute values are equal..$$$$ **Note:** We note that the roots equal when $D=0$ . The roots are distinct when .$D>0,D\ne 0$The roots are rational when $D$ is a perfect square. The roots are integral and distinct when $D>0,D\ne 0$, $D$ is a perfect square and $2a$ divides $D$ exactly.