Question

How do you find the solutions of $- 5 + 2{x^2} = - 6x$?

Answer 1

Write the equation in a decreasing order of the power of variable $x$ to convert the equation in standard form of quadratic equation and then use the quadratic formula. The roots or solutions of a quadratic equation in a standard form $a{x^2} + bx + c = 0$ can be calculated by the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step-by-step solution:
The given equation is $- 5 + 2{x^2} = - 6x$.
Write the equation in a decreasing order of the power of variable $x$ as shown below.
$2{x^2} + 6x - 5 = 0$ …… (1)
It is observed that the highest power of the variable $x$ is $2$ therefore the given equation is a quadratic equation.
The standard form of the quadratic equation is $a{x^2} + bx + c = 0$. Compare the given quadratic equation (1) with the standard quadratic equation and obtain coefficients $a$, $b$ and $c$ as shown below.
$\left( 2 \right){x^2} + \left( 6 \right)x + \left( { - 5} \right) = 0$
It is observed that $a = 2$, $b = 6$ and $c = - 5$.
Now, use the quadratic formula to obtain the solution for the given equation as follows:
$x = \dfrac{{ - \left( 6 \right) \pm \sqrt {{{\left( 6 \right)}^2} - 4\left( 2 \right)\left( { - 5} \right)} }}{{2\left( 2 \right)}}$
Simplify the fraction as shown below.
$\Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 + 40} }}{4}$
$\Rightarrow x = \dfrac{{ - 6 \pm \sqrt {76} }}{4}$
$\Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt {19} }}{4}$
Further simplify the solution as shown below.
$\Rightarrow x = \dfrac{{ - 6}}{4} \pm \dfrac{{2\sqrt {19} }}{4}$
$\Rightarrow x = \dfrac{{ - 3}}{2} \pm \dfrac{{\sqrt {19} }}{2}$
Thus, two solutions of the given quadratic equation $- 5 + 2{x^2} = - 6x$ are $x = \dfrac{{ - 3}}{2} + \dfrac{{\sqrt {19} }}{2}$ and $x = \dfrac{{ - 3}}{2} - \dfrac{{\sqrt {19} }}{2}$.

Note: For a quadratic equation $a{x^2} + bx + c = 0$, two solutions are possible. Type of a solution for a quadratic equation depends on the value of the discriminant $D = \sqrt {{b^2} - 4ac}$. If discriminant is less than zero than two solutions are complex in nature, if discriminant is greater than zero than two solutions are real and distinct in nature and if the discriminant is equal to zero than two solutions are real and same in nature.
Also if we know two solutions (say $\alpha$ and $\beta$) for a quadratic equation then the quadratic equation for this pair of solution is ${x^2} - \left( {\alpha + \beta } \right)x + \left( {\alpha \beta } \right) = 0$.
Generally, the middle term splitting method is not useful to solve a quadratic equation as solutions are sometimes in fraction or irrational number therefore use quadratic formula as a first weapon to tackle quadratic equation.