Question

2x^2+ √5x -5 =0
Use quadratic formula to solve this question

Answer 1

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

$\green{\tt{\therefore{Value\:of\:x\approx 1.1\:and\:-2.2}}}$

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

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

• According to given question :

$\tt \circ \: {2x}^{2} + \sqrt{5} x - 5 = 0 \\ \\ \tt \circ \: a = 2 \\ \\ \tt \circ \: b= \sqrt{5} \\ \\ \tt \circ \: c = - 5 \\ \\ \bold{As \: we \: know \: that}$

$\tt: \implies D = {b}^{2} - 4ac \\ \\ \tt: \implies D= {( \sqrt{5} )}^{2} - 4 \times 2 \times ( - 5) \\ \\ \tt: \implies D=5 + 40 \\ \\ \green{\tt: \implies D=45} \\ \\ \tt: \implies x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \tt: \implies x = \frac{ - \sqrt{5} \pm \sqrt{45} }{2 \times 2} \\ \\ \tt: \implies x = \frac{ - \sqrt{5} \pm \sqrt{45} }{4} \\ \\ \tt \circ \: \sqrt{5} = 2.23 \\ \\ \tt \circ \: \sqrt{45} = 6.7 \\ \\ \tt: \implies x = \frac{ - 2.23 \pm6.7}{4} \\ \\ \green{\tt: \implies x = 1.1175 \: and - 2.2325}$

Answer 2

Given polynomial:-

$\bold{2x^2+\sqrt{5}x-5=0}$

To find:-

The zeros of the polynomial using the quadratic formula.

Solution:-

In the given polynomial 2x² + 5√x - 5 = 0,

• a(coefficient of x²) = 2

• b(coefficient of x) = √5

• c(constant term) = -5

We know the quadratic formula:

$\boxed{x=\frac{-b\±\sqrt{D} }{2a} }$

Now, before that,

D = b²-4ac

⇒D = (√5)²-4*2*(-5)

⇒D = 5 + 40

⇒D = 45

Now, operating for the value of 'x':

$\bold{x=\frac{-\sqrt{5}\±\sqrt{45} }{2*2} }$

⇒ $\bold{x=\frac{-\sqrt{5}\±\sqrt{45} }{4} }$

⇒ $\bold{x=\frac{-\sqrt{5}\±3\sqrt{5} }{4} }$

⇒ $\bold{x=\frac{-\sqrt{5}+3\sqrt{5} }{4} }\:\:or\:\:\:\bold{x=\frac{-\sqrt{5}-3\sqrt{5}}{4} }$

⇒ $\bold{x=\frac{2\sqrt{5} }{4} }\ or\ \bold{x=\frac{-4\sqrt{5} }{4} }$

⇒ $\bold{x=\frac{\sqrt{5}}{2} }\ or\ \bold{x=-\sqrt{5} }$

∴So, the zeros of the quadratic equation are √5/2 or -√5.