Question

Find the zeroes of the quadratic polynomial:
2x2— 5x + 2

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=2\:and\:\frac{1}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {2x}^{2} - 5x + 2 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Value \: of \: x = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\\tt: \implies {2x}^{2} - 5x + 2 = 0 \\ \\ \tt: \implies {2x}^{2} -4x -x + 2 = 0 \\ \\ \tt: \implies 2x(x - 2) - 1(x - 2) = 0 \\ \\ \tt: \implies(2x -1)(x - 2) = 0\\ \\ \green{\tt: \implies x = 2 \: and \: \frac{ 1}{2}} \\ \\ \bold{Alernate \: method} \\ \tt: \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \tt: \implies x = \frac{ - (-5) \pm \sqrt{ {(-5)}^{2} - 4 \times 2 \times 2 } }{2 \times 2} \\ \\ \tt: \implies x = \frac{ 5 \pm \sqrt{9} }{4} \\ \\ \tt: \implies x = \frac{ 5 \pm3}{4} \\ \\ \green{\tt: \implies x = 2 \: and \: \frac{ 1}{2} }$

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$