How do you solve the equation ${{x}^{2}}+169=0?$
Answer
575.7k+ views
Hint: (1) Square root of a negative number is an imaginary number, as, $\sqrt{-4}$ will be $+2i$ and $-2i$, Here, the letter $'i'$ denotes that both the values are imaginary.
(2) Convert the given equation in form of ${{x}^{2}}=c$ then determine the value of $x.$
Complete step-by-step solution:
As per data given in the question,
We have,
${{x}^{2}}+169=0$
As the above equation has $2$ in power of $x,$ so, it is a type of quadratic equation.
We know that,
If we add or subtract number from both side of equation, then the value of equation doesn’t changes.
So,
Subtracting $169$ from both side of equation,
We will get,
${{x}^{2}}+169=0$
$\Rightarrow {{x}^{2}}+169-169=-169$
Here, in LHS, $169$ and $-169$ get cancelled,
So, equation becomes,
${{x}^{2}}=-169$
$\Rightarrow x=\sqrt{-169}...(i)$
As we know that,
Square root of negative number is an imaginary number and
As,
${{\left( x \right)}^{2}}=\left( x.x \right)={{x}^{2}}$ and ${{\left( -x \right)}^{2}}=\left( -x \right)\left( -x \right)={{x}^{2}}$
It means, square root of ${{x}^{2}}$ will be both $\left( +x \right)$ and $\left( -x \right)$ applying both concept in equation $\left( i \right)$
As,
$x=\sqrt{-169}$
$\Rightarrow x=-\sqrt{169}$
$\Rightarrow x=-\sqrt{{{13}^{2}}}$
$\Rightarrow x=+13i,-13i$
So, value of equation,
${{x}^{2}}+169=0$ will be $+13i$ and $-13i$
Additional Information: (1) As, ${{x}^{2}}+169=0$
Here, compare the given equation with the general equation of the quadratic equation.
The general form of the quadratic equation is. $a{{x}^{2}}+bx+c=0$
Where, $a,b$ and $c$ are constant.
So, roots can also be determined by,
$=\dfrac{-b\pm \sqrt{d}}{2a}$
Where $D$ is called as determinant
Value of $D={{b}^{2}}-4ac$
Hence we can also determine the solution of a given equation by using the above mentioned formula.
Note: (1) Square root of positive number is always a real number like $\sqrt{9}=+3,-3$
(2) Square root of negative number is always an imaginary number like, $\sqrt{9}=+3i,-3i$
(3) While adding and subtracting or doing any mathematical operation make sure you are doing the operations with the same number in both the left hand and right hand side of the equation.
(2) Convert the given equation in form of ${{x}^{2}}=c$ then determine the value of $x.$
Complete step-by-step solution:
As per data given in the question,
We have,
${{x}^{2}}+169=0$
As the above equation has $2$ in power of $x,$ so, it is a type of quadratic equation.
We know that,
If we add or subtract number from both side of equation, then the value of equation doesn’t changes.
So,
Subtracting $169$ from both side of equation,
We will get,
${{x}^{2}}+169=0$
$\Rightarrow {{x}^{2}}+169-169=-169$
Here, in LHS, $169$ and $-169$ get cancelled,
So, equation becomes,
${{x}^{2}}=-169$
$\Rightarrow x=\sqrt{-169}...(i)$
As we know that,
Square root of negative number is an imaginary number and
As,
${{\left( x \right)}^{2}}=\left( x.x \right)={{x}^{2}}$ and ${{\left( -x \right)}^{2}}=\left( -x \right)\left( -x \right)={{x}^{2}}$
It means, square root of ${{x}^{2}}$ will be both $\left( +x \right)$ and $\left( -x \right)$ applying both concept in equation $\left( i \right)$
As,
$x=\sqrt{-169}$
$\Rightarrow x=-\sqrt{169}$
$\Rightarrow x=-\sqrt{{{13}^{2}}}$
$\Rightarrow x=+13i,-13i$
So, value of equation,
${{x}^{2}}+169=0$ will be $+13i$ and $-13i$
Additional Information: (1) As, ${{x}^{2}}+169=0$
Here, compare the given equation with the general equation of the quadratic equation.
The general form of the quadratic equation is. $a{{x}^{2}}+bx+c=0$
Where, $a,b$ and $c$ are constant.
So, roots can also be determined by,
$=\dfrac{-b\pm \sqrt{d}}{2a}$
Where $D$ is called as determinant
Value of $D={{b}^{2}}-4ac$
Hence we can also determine the solution of a given equation by using the above mentioned formula.
Note: (1) Square root of positive number is always a real number like $\sqrt{9}=+3,-3$
(2) Square root of negative number is always an imaginary number like, $\sqrt{9}=+3i,-3i$
(3) While adding and subtracting or doing any mathematical operation make sure you are doing the operations with the same number in both the left hand and right hand side of the equation.
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