Question

Factorise the expression \[2{x^2} + 5x - 3 = 0\].

Answer 1

Hint: Factorisation is the process in which an expression is broken up into numbers which in turn can be multiplied to get the original expression.

Complete step by step solution:
For any quadratic equation we always have two probable solutions for \[x\]. To factorise the above quadratic try to find any two numbers which when multiplied will be equal to the constant value \[ac\], and when subtracted (in this case, or added if the sign before the constant \[c\] is ‘\[ + \]’) will be equal to \[b\].
Here in this sum \[a = 2,\] \[b = 5\], \[c = 3\], \[ac = 6\] . Factorise \[6\] and try to find the required number. Observe that \[6\] when multiplied by \[1\] equals \[6\]and \[1\] when subtracted from \[6\] gives \[5\] that is equal to \[b\].
Hence both the numbers are \[1\] and \[6\]. Now express \[b\] as the difference of the two numbers as shown:
\[\therefore \] \[2{x^2} + 5x - 3\] \[ = \] \[2{x^2} + 6x - x - 3\]
Take out the common factor from the first two terms and last two terms respectively:
\[ = \] \[2x\left( {x + 3} \right) - 1\left( {x + 3} \right)\]
Again take out the common factor:
\[ = \] \[\left( {x + 3} \right)\left( {2x - 1} \right)\]
To find out the factors, we equate \[\left( {x + 3} \right) = 0\] and \[\left( {2x - 1} \right) = 0\]
\[ \Rightarrow \left( {2x - 1} \right) = 0\] or \[\left( {x + 3} \right) = 0\]
\[ \Rightarrow x = \dfrac{1}{2}\] or \[x = - 3\]

Additional Information:
Any equation of the form \[a{x^2} + bx + c = 0\], \[a \ne 0\], where \[a,b,c\] are constants and \[x\] is a variable is known as a quadratic equation.
Here, \[{b^2} - 4ac\] is called the discriminant \[D\], of the quadratic equation.
For any quadratic equation:
If \[D > 0\], roots are real and unequal.
If \[D = 0\], roots are real and equal.
If \[D < 0\], roots are imaginary.

Note: Be careful about the sign in form of the constant \[c\]. Sometimes the product of \[ac\] may have some factors which when added gives \[b\] and there may be two factors which when subtracted also gives\[b\]. In that case the numbers must be chosen based on the operation that needs to be performed depending upon the plus or minus sign before. If it is a \[' + '\] sign then we choose the numbers which when added gives \[b\], if it is \[' - '\] then we choose the numbers which when subtracted gives \[b\].