Question

factor and then solve equation 2x^2-5x-25=0​

Answer 1

Please see the description image

Answer 2

Question

$\rm \: \implies2x {}^{2} - 5x - 25 = 0$

Solution:-

Method:- 1 factorization method

$\rm \: \implies2x {}^{2} - 5x - 25 = 0$

Splitting into middle term

$\rm \: \implies2x {}^{2} - 10x + 5x - 25 = 0$

$\implies \rm \: 2x(x - 5) + 5(x - 5) = 0$

$\rm \: \implies(2x + 5)(x - 5) = 0$

$\rm \: x = - \dfrac{5}{2} \: \: and \: \: x = 5$

Method :- 2 quadratic formula

$\rm \: \implies2x {}^{2} - 5x - 25 = 0$

Compare with:- ax² + bx + c = 0

$\rm \: a = 2,b = - 5 \: \: and \: \: c = - 25$

Quadratic formula

$\rm \: x = \dfrac{ -b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$\rm \: x = \dfrac{ - ( - 5) \pm \sqrt{( - 5) {}^{2} - 4 \times 2 \times - 25 } }{2 \times 2}$

$\rm \: x = \dfrac{5 \pm \sqrt{25 + 200} }{4}$

$\rm \: x = \dfrac{5 \pm \: \sqrt{225} }{4}$

$\rm \: x = \dfrac{5 \pm15}{4}$

$\rm \: x = \dfrac{15 + 5}{4} \: and \: \dfrac{ - 15 + 5}{4}$

$\rm \: x \: = \dfrac{20}{4} \: \: and \: \: \dfrac{ - 10}{4}$

$\rm \: x = 5 \: \: \: and \: \: \dfrac{ - 5}{2}$