Question

How do you solve $2{x^2} - 5x = - 7$ using the quadratic formula?

Answer 1

First move $7$ to the left side of the equation by adding $7$ to both sides of the equation. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the 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.

Formula used:
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 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

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 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first move $7$ to the left side of the equation by adding $7$ to both sides of the equation.
$2{x^2} - 5x + 7 = 0$
Next, compare $2{x^2} - 5x + 7 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $2{x^2} - 5x + 7 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 2$, $b = - 5$ and $c = 7$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 5} \right)^2} - 4\left( 2 \right)\left( 7 \right)$
After simplifying the result, we get
$\Rightarrow D = 25 - 56$
$\Rightarrow D = - 31$
Which means the given equation has no real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {31} i}}{{2 \times 2}}$
It can be written as
$\Rightarrow x = \dfrac{{5 \pm \sqrt {31} i}}{4}$
$\Rightarrow x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ and $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$
So, $x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ and $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$ are roots/solutions of equation $2{x^2} - 5x = - 7$.

Therefore, the solutions to the quadratic equation $2{x^2} - 5x = - 7$ are $x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ and $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$.

Note: We can check whether $x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ and $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$ are roots/solutions of equation $2{x^2} - 5x = - 7$ by putting the value of $x$ in given equation.
Putting $x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ in LHS of equation $2{x^2} - 5x = - 7$.
${\text{LHS}} = 2{\left( {\dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i} \right)^2} - 5\left( {\dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i} \right)$
On simplification, we get
$\Rightarrow {\text{LHS}} = 2\left( {\dfrac{{25}}{{16}} - \dfrac{{31}}{{16}} + \dfrac{{5\sqrt {31} }}{8}i} \right) - \dfrac{{25}}{4} - \dfrac{{5\sqrt {31} }}{4}i$
$\Rightarrow {\text{LHS}} = - \dfrac{3}{4} + \dfrac{{5\sqrt {31} }}{4}i - \dfrac{{25}}{4} - \dfrac{{5\sqrt {31} }}{4}i$
$\Rightarrow {\text{LHS}} = - 7$
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ is a solution of equation $2{x^2} - 5x = - 7$.
Putting $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$ in LHS of equation $2{x^2} - 5x = - 7$.
${\text{LHS}} = 2{\left( {\dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i} \right)^2} - 5\left( {\dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i} \right)$
On simplification, we get
$\Rightarrow {\text{LHS}} = 2\left( {\dfrac{{25}}{{16}} - \dfrac{{31}}{{16}} - \dfrac{{5\sqrt {31} }}{8}i} \right) - \dfrac{{25}}{4} + \dfrac{{5\sqrt {31} }}{4}i$
$\Rightarrow {\text{LHS}} = - \dfrac{3}{4} - \dfrac{{5\sqrt {31} }}{4}i - \dfrac{{25}}{4} + \dfrac{{5\sqrt {31} }}{4}i$
$\Rightarrow {\text{LHS}} = - 7$
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$ is a solution of equation $2{x^2} - 5x = - 7$.
Final solution: Therefore, the solutions to the quadratic equation $2{x^2} - 5x = - 7$ are $x = \dfrac{5}{4} + \dfrac{{\sqrt {31} }}{4}i$ and $x = \dfrac{5}{4} - \dfrac{{\sqrt {31} }}{4}i$.