Question

How do you solve $4{x^2} + 5x = 6$?

Answer 1

Hint: In the above question, we are asked to solve a quadratic equation. A quadratic equation always has 2 values. To solve a quadratic equation, we need to equate the given equation with the standard quadratic equation, which, further has a standard quadratic formula, where the corresponding values are to be substituted, to find the desired values.

Complete step by step solution:
The given equation here is: $4{x^2} + 5x = 6$
First, transpose 6 to the other side so the equation would end up in the standard form: \[a{x^2} + bx + c = 0\]
Thus, we get:
$4{x^2} + 5x - 6 = 0$ ……………………….(1)
Now, we have a quadratic equation whose roots (here, ‘x’) are to be determined.
We know that, the roots of the quadratic equation$A{x^2} + Bx + C = 0$ are given by the quadratic formula: $\dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Thus, on comparing equation (1) with $A{x^2} + Bx + C = 0$, we get: $A = 4,B = 5,C = - 6$
Hence, the roots are given as:
$
  x = \dfrac{{ - 5 \pm \sqrt {{5^2} - 4\left( 4 \right)\left( { - 6} \right)} }}{{2\left( 4 \right)}} \\
   = \dfrac{{ - 5 \pm \sqrt {25 + 96} }}{8} \\
   = \dfrac{{ - 5 \pm \sqrt {121} }}{8} \\
 $
Again, the square root of 121 is 11.
Hence, the roots are:
$x = \dfrac{{ - 5 \pm 11}}{8}$
Which can further be written as:
$x = \dfrac{{ - 5 + 11}}{8} = \dfrac{6}{8} = \dfrac{3}{4}$ and $x = \dfrac{{ - 5 - 11}}{8} = \dfrac{{ - 16}}{8} = - 2$ respectively.
Thus, the two distinctive solutions of the equation $4{x^2} + 5x = 6$ are given as:
$x = \dfrac{3}{4}$ and $x = 2$


Note: A quadratic equation is one which is of the form:$A{x^2} + Bx + C = 0$, where A, B, C are real numbers and \[A \ne 0\]. The coefficient of ${x^2}$, ‘A’, is called the quadratic coefficient. The coefficient of $x$, ‘B’, is called the linear coefficient. While, ‘C’ is the constant term. The value of ‘x’ satisfying the above quadratic equation are the roots of the equation $\left( {\alpha ,\beta } \right)$. ‘x’ will always have two values, i.e., there will be 2 roots of the quadratic equation, the nature of which may be either real or imaginary. Nature of the roots are determined by the values of the discriminant(D), given by: $D = {B^2} - 4AC$. The roots are:
i) Equal, if $D = 0$
ii) Imaginary, if D < 0
iii) Real, if D > 0
iv) Rational, if D > 0 and ‘D’ is a perfect square
v) Irrational, if D > 0 and ‘D’ is not a perfect square
vi) Integral, if D > 0, ‘D’ is a perfect square, A = 1, and B & C are integers.