Question

2xsq-7x-39 factorise and find zeros of the polynomial

Answer 1

Given :-

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

What to find out =Roots of the equation?

Solution :-

Factorise by splitting middle term

$2 {x}^{2} - 7x - 39 = 0 \\ \\ 2 {x}^{2} + 6x - 13x - 39 = 0 \\ \\ 2x(x + 3) - 13(x + 3) = 0 \\ \\ (x + 3)(2x - 13) = 0$

Now split it into possible cases

$(2x - 13) = 0 \\ \\ (x + 3) = 0$

Hence,

$x_1 = ( \frac{13}{2} ) \\ \\ x_2 = (3)$

Extra information :-

• An equation containing a single variable of degree 2. Its general form is ax^2 + bx + c = 0, where x is the variable and a, b, and c are constants (a ≠ 0).

Answer 2

ANSWER:-

➣

• Given:-

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

• To Find:-

Roots of the given equation.

Solution :-

• Factorization by Splitting Middle term:-

$2 {x}^{2} - 7x - 39 = 0 \\ \\ \implies2 {x}^{2} + 6x - 13x - 39 = 0 \\ \\ \implies2x(x + 3) - 13(x + 3) = 0 \\ \\ \implies(x + 3)(2x - 13) = 0$

• Next step is to Split into possible cases:-

$(2x - 13) = 0$

$(x + 3) = 0$

• Taking 1st one:-

$2x - 13 = 0$

$2x = 13$

$x = {\frac{13}{2}}$

• Now, 2nd one:-

$x + 3 = 0$

$x = -3$

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Therefore, we have::

$\bf{x_1 = \frac{13}{2} }$

$\bf{x_2 = 3}$