Question

How do you solve \[6{x^2} + 48x - 54 = 0\]?

Answer 1

Hint: Here in this question, we have to solve the given equation, the given equation is in the form of a quadratic equation. This is a quadratic equation for the variable x. By using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we can determine the solutions.

Complete step-by-step solution:
The question involves the quadratic equation. To the quadratic equation we can find the roots by factoring or by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. consider the given equation \[6{x^2} + 48x - 54 = 0\].
In general, the quadratic equation is represented as \[a{x^2} + bx + c = 0\], when we compare the above equation to the general form of equation the values are as follows. a=6 b=48 and c=-54. Now substituting these values to the formula for obtaining the roots we have
\[\Rightarrow x = \dfrac{{ - (48) \pm \sqrt {{{(48)}^2} - 4(6)( - 54)} }}{{2(6)}}\]
On simplifying the terms, we have
\[ \Rightarrow x = \dfrac{{ - 48 \pm \sqrt {2304 + 1296} }}{{12}}\]
Now add 2304 to 1296 we get
\[ \Rightarrow x = \dfrac{{ - 48 \pm \sqrt {3600} }}{{12}}\]
The number 3600 is a perfect square number and we have a square root for this. So, the square root of the number 3600 is 60 so we have.
Therefore, we have \[x = \dfrac{{ - 48 + 60}}{{12}}\] or \[x = \dfrac{{ - 48 - 60}}{{12}}\]. We can simplify for further so we get
\[ \Rightarrow x = \dfrac{{12}}{{12}}\] and \[x = \dfrac{{ - 108}}{{12}}\]
So we have
\[ \Rightarrow x = 1\] and \[x = - 9\]
hence we have solved the quadratic equation and found the value of the variable x.
The equation is also solved by using the factorisation method.

Note: The quadratic equation can be solved by using the factorisation method and we also find the roots by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. While factorising we use sum product rule, the sum product rule is given as the product factors of the number c is equal to the sum of the factors which satisfies the value of b.