Question

Solve the equation $4{{x}^{2}}-25=0$.

Answer 1

Hint: In this question, we are given a quadratic equation and we have to solve it i.e. we need to find a value of x which satisfies the given equation. For this, we will use the factorization method. Using ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we will form factors of the above equation and then take them equal to 0 and solve for x. Since the equation is of degree two, so we will have two values of x.

Complete step-by-step answer:
Here, we are given the expression as $4{{x}^{2}}-25=0$. Let us solve it using the factorization method.
We know 4 can be written as ${{\left( 2 \right)}^{2}}$. So, we can write $4{{x}^{2}}$ as ${{\left( 2x \right)}^{2}}$. Also, we know $5\times 5=25\Rightarrow {{5}^{2}}=25$. Hence, we can write 25 as ${{5}^{2}}$.
So now our equation becomes ${{\left( 2x \right)}^{2}}-{{\left( 5 \right)}^{2}}=0$.
Since, both terms are perfect squares, so let us factorize it using the difference of squares formula given by ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Hence, a will be equal to 2x and b will be equal to 5, hence we get:
$\left( 2x+5 \right)\left( 2x-5 \right)=0$.
As we know, if any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
Therefore, $2x+5=0\text{ and }2x-5=0$.
Let us solve first factor,
$2x+5=0\Rightarrow 2x=-5$.
Dividing by 2 both sides we get:
$x=\dfrac{-5}{2}$.
Now solving second factor,
$2x-5=0\Rightarrow 2x=5$.
Dividing by 2 both sides, we get:
$x=\dfrac{5}{2}$.
Hence, $x=\dfrac{-5}{2},\dfrac{5}{2}$ are required values.

Note: Students can check their answer by putting value in the given equation and verifying if the left side of the equation is equal to the right side of the equation. Students can also use quadratic formulas for solving this equation. Quadratic formula for an equation $a{{x}^{2}}+bx+c$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In $4{{x}^{2}}-25$, a = 4 and c = -25. So,
\[\begin{align}
  & x=\dfrac{0\pm \sqrt{0-4\left( 4 \right)\left( 25 \right)}}{2\left( 4 \right)} \\
 & \Rightarrow x=\dfrac{\pm \sqrt{400}}{8} \\
 & \Rightarrow x=\dfrac{\pm 20}{8} \\
 & \Rightarrow x=\dfrac{+20}{8},x=\dfrac{-20}{8} \\
 & \Rightarrow x=\dfrac{5}{2},x=\dfrac{-5}{2} \\
\end{align}\]
Which is the same solution as found by the factorization method. Students should take care of the signs while solving the equation.