How do you solve ${x^2} - 20x + 99 = 0$?
Answer
613.8k+ views
Hint: First compare the given quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.
Complete step by step answer:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
First, compare ${x^2} - 20x + 99 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - 20x + 99 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 20$ and $c = 99$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 20} \right)^2} - 4\left( 1 \right)\left( {99} \right)$
After simplifying the result, we get
$ \Rightarrow D = 400 - 396$
$ \Rightarrow D = 4$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \left( { - 20} \right) \pm 2}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = 10 \pm 1$
$ \Rightarrow x = 11$ and $x = 9$
So, $x = 9$ and $x = 11$ are roots/solutions of equation ${x^2} - 20x + 99 = 0$.
Note: Note that quadratic formula is used to find the roots of a given quadratic equation. Sometimes we can factorize the roots directly. But quadratic formulas can be used generally to find the roots of any quadratic equation. The value of discriminant is used to decide the type of roots so obtained such that roots are equal or different and are real or not.
Complete step by step answer:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
First, compare ${x^2} - 20x + 99 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - 20x + 99 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 20$ and $c = 99$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 20} \right)^2} - 4\left( 1 \right)\left( {99} \right)$
After simplifying the result, we get
$ \Rightarrow D = 400 - 396$
$ \Rightarrow D = 4$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \left( { - 20} \right) \pm 2}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = 10 \pm 1$
$ \Rightarrow x = 11$ and $x = 9$
So, $x = 9$ and $x = 11$ are roots/solutions of equation ${x^2} - 20x + 99 = 0$.
Note: Note that quadratic formula is used to find the roots of a given quadratic equation. Sometimes we can factorize the roots directly. But quadratic formulas can be used generally to find the roots of any quadratic equation. The value of discriminant is used to decide the type of roots so obtained such that roots are equal or different and are real or not.
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