Question

$2x^{2} -7x-39=0$

Answer 1

Answer:

$2 {x}^{2} - 7x - 39 = 0 \\ 2 {x }^{2} - (13 - 6)x - 39 = 0 \\ 2 {x }^{2} - 13x + 6x - 39 = 0 \\ 2x(x + 3) - 13(x + 3) = 0 \\( 2x - 13) \times (x + 3) = 0 \\ therefore \\ either \\ \: 2x - 13 = 0 \\ 2x = 13 \\ x = \frac{13}{2} \\ x = 6.5 \\ or \\ x + 3 = 0 \\ x = - 3$

Step-by-step explanation:

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Answer 2

To factorize : $2x^{2} -7x -39 = 0$

Solution :

• According to question we are asked to split the given expression by splitting the middle term.
• We spilt the middle term such that product of the numbers is last term and sum or difference of those numbers is middle term.
• Here, $-7x$ is the middle term and on multiplying constants of first term and last we get $78$ as  the last term.
• $2x^{2} -7x -39 = 0$

      $= 2x^{2} -13x + 6x -39 \\ =x (2x -13 ) + 3 (2x - 13)\\=( 2x -13 ) (x + 3 )$

• Therefore , $x = \frac{13}{2}$  or $x = -3$

Hence, on splitting the middle term of $2x^{2} -7x -39 = 0$ we get value of $x$ as $x = \frac{13}{2}$  or $x = -3$ .