Question

How do solve \[{{x}^{2}}+3x-180=0\]?

Answer 1

Hint: This type of problem is based on the concept of solving quadratic equations. First, we have to find the discriminant \[\Delta \] of the given equation. We find that \[\Delta >0\]. Therefore, the two roots are real and distinct. Then we need to find a pair of terms \[\left( a,b \right)\] which satisfies the conditions a + b = 3 and \[ab=-180\]. Then, convert the given equation as \[{{x}^{2}}-\left( a+b \right)x+ab=0\]. Expand the obtained equation and factorize. We then make the necessary calculations to obtain the required solution of the given equation.

Complete step by step answer:
According to the question, we are asked to solve the given equation \[{{x}^{2}}+3x-180=0\].
We have given the equation is \[{{x}^{2}}+3x-180=0\] -----(1)
We first have to find the discriminant \[\Delta \].
And we know that
\[\Delta ={{b}^{2}}-4ac\] for the quadratic equation \[a{{x}^{2}}+bx+c=0\].
And if we compare \[{{x}^{2}}+3x-180=0\] and \[a{{x}^{2}}+bx+c=0\], we get a=1, b=3 and c=-180.
\[\Rightarrow \Delta ={{3}^{2}}-4\left( 1 \right)\left( -180 \right)\]
On further simplification, we get,
\[\Rightarrow \Delta =9+4\left( 180 \right)\]
\[\Rightarrow \Delta =9+720\]
\[\therefore \Delta =729\]
Here we observe that \[\Delta >0\]. Therefore the roots are real and distinct.

Then, we have to find the pair of terms \[\left( a,b \right)\] which satisfies the conditions
 \[a+b=3\] and \[ab=-180\].
Since \[ab<0\], either ‘a’ or ‘b’ is a negative term.
Then the pairs of terms which satisfy \[ab=-180\] are
\[\left( -18,10 \right),\left( -10,18 \right),\left( -15,12 \right),\left( -12,15 \right),\left( -20,9 \right)\] etc.
Now consider \[a+b=3\].
Since \[a+b>0\], we get \[\left| a \right|>\left| b \right|\].
That is, ‘a’ is a positive term and ‘b’ is a negative term.

Therefore, the pair of terms which satisfy \[a+b=3\]and \[ab=-180\] is \[\left( 15,-12 \right)\].
We get a=15 and b= -12.
Now we have to convert equation (1) as \[{{x}^{2}}+\left( a+b \right)x+ab=0\].
\[\Rightarrow {{x}^{2}}+\left( 15-12 \right)x-15\left( 12 \right)=0\]
On further simplification, we get
\[\Rightarrow {{x}^{2}}+15x-12x-15\left( 12 \right)=0\]
Taking x common from the first two terms and -12 common from the last two terms, we get,
\[\Rightarrow x\left( x+15 \right)-12\left( x+15 \right)=0\] -----(2)
Let us now take \[\left( x+15 \right)\] common from the equation (2).
\[\Rightarrow \left( x+15 \right)\left( x-12 \right)=0\] -------(3)
When x.y=0 either x=0 or y=0 or both x,y=0.
Using the above condition in the equation (3), we get,
\[x+15=0\text{ or }x-12=0\]
\[\therefore x=-15\text{ or }x=12\]
Hence, the solutions of the given equation \[{{x}^{2}}+3x-180=0\] are 12 and -15.

Note:
 Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final of x which will be the required answer. We should avoid calculation mistakes based on sign conventions. We can also solve this question by making use of these formula for the value of x in the given equation, that is, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] where a=1, b=3 and c=-180.