
How do you solve using the quadratic formula \[{{x}^{2}}-25=0\]?
Answer
571.5k+ views
Hint: In this given problem we have to solve and find the value x from the given quadratic equation \[{{x}^{2}}-25=0\]. We can solve this equation by two methods, one is the quadratic formula method and the other is the difference of the square formula. In this problem we are asked to solve using the quadratic formula method, so we can solve using the quadratic formula. We can compare the given equation with the general equation to get the value of a, b, c to substitute in the quadratic formula to solve for x.
Complete step by step answer:
We know that the general form of the quadratic equation is,
\[a{{x}^{2}}+bx+c=0\] ……. (1)
We know that the quadratic formula for the quadratic equation is,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] ……. (2)
We also know that the given quadratic equation is,
\[{{x}^{2}}-25=0\] ……. (3)
Now we can compare equation (1) and (3), we get the value of a, b, c.
a = 1, b = 0, c = -25.
Now, we can substitute the above values from the equation in (2).
\[\begin{align}
& \Rightarrow x=\dfrac{-\left( 0 \right)\pm \sqrt{{{\left( 0 \right)}^{2}}-4\left( 1 \right)\left( -25 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{\pm \sqrt{-\left( -100 \right)}}{2} \\
& \Rightarrow x=\dfrac{\pm \sqrt{100}}{2} \\
\end{align}\]
Now, we can simplify the square term to get the value of x.
\[\begin{align}
& \Rightarrow x=\dfrac{\pm 10}{2} \\
& \Rightarrow x=\pm 5 \\
\end{align}\]
Therefore, by solving \[{{x}^{2}}-25=0\] using quadratic formula, \[x=\pm 5\].
Note:
In this problem we used a quadratic formula to solve for x, for this we should know the quadratic formula for the general quadratic equation. We can also use the difference of the squares to solve for x, we can also check values by using the difference of square formula.
Complete step by step answer:
We know that the general form of the quadratic equation is,
\[a{{x}^{2}}+bx+c=0\] ……. (1)
We know that the quadratic formula for the quadratic equation is,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] ……. (2)
We also know that the given quadratic equation is,
\[{{x}^{2}}-25=0\] ……. (3)
Now we can compare equation (1) and (3), we get the value of a, b, c.
a = 1, b = 0, c = -25.
Now, we can substitute the above values from the equation in (2).
\[\begin{align}
& \Rightarrow x=\dfrac{-\left( 0 \right)\pm \sqrt{{{\left( 0 \right)}^{2}}-4\left( 1 \right)\left( -25 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{\pm \sqrt{-\left( -100 \right)}}{2} \\
& \Rightarrow x=\dfrac{\pm \sqrt{100}}{2} \\
\end{align}\]
Now, we can simplify the square term to get the value of x.
\[\begin{align}
& \Rightarrow x=\dfrac{\pm 10}{2} \\
& \Rightarrow x=\pm 5 \\
\end{align}\]
Therefore, by solving \[{{x}^{2}}-25=0\] using quadratic formula, \[x=\pm 5\].
Note:
In this problem we used a quadratic formula to solve for x, for this we should know the quadratic formula for the general quadratic equation. We can also use the difference of the squares to solve for x, we can also check values by using the difference of square formula.
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