Question

Solve the given equation to find the values of $x$: ${x^2} - 8x + 15$$ = 0$

Answer 1

Hint: We have to solve the given quadratic equation to find the value of $x$. For that, first we have to factorize the quadratic equation. Always make sure that the quadratic equation is in the $a{x^2} + bx + c$$ = 0$ format. Compare the terms of the given equation with the standard format and use it in the solution. Here we factorize by using the property of relation between roots and coefficient. Finally we will get the solution.

Formula used: Property of relationship between roots and coefficient,
$x + y = - \dfrac{b}{a}$ and
$x \cdot y = \dfrac{c}{a}$

Complete step-by-step solution:
The given quadratic equation is,
 ${x^2} - 8x + 15 = 0$
Now comparing it to the standard equation, $a{x^2} + bx + c$, we get:
$a = 1$, $b = - 8$, and $c = 15$
Now, we can solve the given quadratic equation using two methods.
Factoring – by using the factorization method.
Let $x$ and $y$ be the roots of the equation,
We are going to use the property of relationship between roots and coefficient,
$x + y = - \dfrac{{ - 8}}{1} = 8$
$x \cdot y = \dfrac{{15}}{1} = 15$
The condition is, Now if we multiply the factors, it should give $a$, and if we either add or subtract, it should give $b$.
In the given equation, $x \cdot y = 15$
Factors of $15 = 1,{\text{ }}3,{\text{ }}5,{\text{ }}15$
If we multiply $3,5$ we get $15$ and add $3,5$ we get $8$.
Splitting the middle term,
${x^2} - 3x - 5x + 15 = 0$
Simplifying we get,
$ \Rightarrow x(x - 3) - 5(x - 3)$
Hence,
$ \Rightarrow (x - 3)(x - 5)$
Now we equate $(x - 3)(x - 5) = 0$
Case 1:
Let us consider $\left( {x - 3} \right) = 0$
$ \Rightarrow x = 3$
Case 2:
Let us consider $\left( {x - 5} \right) = 0$
$ \Rightarrow x = 5$
Here we get,
$\therefore $ $x = 3$ and $y = 5$

The roots of the equation are 3 and 5.

Note: Quadratic formula – by using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The given quadratic equation is,
 ${x^2} - 8x + 15 = 0$
Now comparing it to the standard equation, $a{x^2} + bx + c$, we get:
$a = 1$, $b = - 8$, and $c = 15$
Going with the quadratic formula and substituting the given $a,b,c$ values in:
$ \Rightarrow \dfrac{{ - ( - 8) \pm \sqrt {{{( - 8)}^2} - 4(1)(15)} }}{{2(1)}}$
Simplifying we get,
$ \Rightarrow \dfrac{{8 \pm \sqrt {64 - 60} }}{2}$
Subtracting the term,
$ \Rightarrow \dfrac{{8 \pm \sqrt 4 }}{2}$
Taking square root we get,
$ \Rightarrow \dfrac{{8 \pm 2}}{2} - - - ( * )$
Now, we have two cases from equation (*)
Case 1:
Let us consider the term $\dfrac{{8 + 2}}{2}$
Hence one root is,
$ \Rightarrow \dfrac{{10}}{2} = 5$
Case 2:
Let us consider the term $\dfrac{{8 - 2}}{2}$
Hence one root is,
$ \Rightarrow \dfrac{6}{2} = 3$
Here, we have two values for $x$:
$\therefore $ $x = 3$ and $y = 5$
We can solve the same quadratic equation with the factorization method. Solve the equation using that method for more practice and sharp calculations and the cross check with the given solution. You can also use the square root method to obtain the answers. Foe better understanding, find the discriminant of the given equation to know if the solution or the roots are real or imaginary or equal. If they are proved to be imaginary, the only way to find the roots is through the quadratic equation method.