Question

factorise : (1).6x2-3-7x​

Answer 1

Product of coefficient of x^2 and constant term = 6*-3 = - 18

Rearranging on the basis of power of x,

6x^2-7x-3

6x^2-9x+2x-3

3x(2x-3) +1(2x-3)

(3x+1)(2x-3)

Answer 2

Question :-

$6 {x}^{2} - 7x - 3 = 0$

What to find out =Roots of the equation?

Solution :-

Factorise by splitting middle term

$6 {x}^{2} - 7x - 3 = 0 \\ \\ 6 {x}^{2} - 9x + 2x - 3 = 0 \\ \\ 3x(2x - 3) + 1(2x - 3) = 0 \\ \\ (2x - 3)(3x + 1) = 0$

Now split it into possible cases

$(2x - 3) = 0 \\ \\ (3x + 1) = 0$

Hence,

$x_1 = ( \frac{3}{2} ) \\ \\ x_2 = ( - \frac{1}{3} )$

Extra information :-

• A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.