Question

How do you solve $8x - 10 = {x^2} - 7x + 3$ using the quadratic formula?

Answer 1

We first need to rearrange the given equation in the standard form of quadratic equation. That is in the form of $a{x^2} + bx + c = 0$. After that we can solve this using various methods that are by completing the square, factorization, graph or by quadratic formula. Here we need to use a quadratic formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step-by-step solution:
Given, $8x - 10 = {x^2} - 7x + 3$
Shifting the terms we have,
${x^2} - 7x + 3 - 8x + 10 = 0$
${x^2} - 15x + 13 = 0$
Since the degree of the equation is 2, we have 2 factors or two roots.
On comparing the given equation with the standard quadratic equation $a{x^2} + bx + c = 0$, we have $a = 1$, $b = - 15$ and $c = 13$.
Now we use quadratic formula,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Substituting we have,
$\Rightarrow x = \dfrac{{ - ( - 15) \pm \sqrt {{{\left( {15} \right)}^2} - 4(1)(13)} }}{{2(1)}}$
$= \dfrac{{15 \pm \sqrt {225 - 52} }}{2}$
$= \dfrac{{15 \pm \sqrt {173} }}{2}$
Thus we have two roots,
$\Rightarrow x = \dfrac{{15 + \sqrt {173} }}{2}$ and $x = \dfrac{{15 - \sqrt {173} }}{2}$. This is the required answer.

Note: Since we have a polynomial of degree two and hence it is called quadratic polynomial. If we have a polynomial of degree ‘n’ then we have ‘n’ roots. In the given problem we have a degree that is equal to 2. Hence the number of roots are 2. We use a quadratic formula if we are unable to split the middle term that fails to factorize. Quadratic formula and Sridhar’s formula are both the same. We know that the product of two negative numbers gives us a positive number. Also keep in mind when shifting values from one side of the equation to another side of the equation, always change sign from positive to negative and vice-versa.