Question

if the real root exists of the following quadratic equation 2x^2-5x-3=0 then find the roots of the equations
​

Answer 1

Answer:

To find -Root and Discriminant

Step-by-step explanation:

a=2,b=-5,c=-3

D=b^2-4ac

=>(-5)^2-4*2*(-3)

=>25+24

=>49

This equation will have roots because D>0

x=-b/2a+-√49/2a

=>5/4+-7/4

=>5/4+7/4 or 5/4-7/4

=>12/4 or -2/4

=>x=3,-1/2

Hope this will help you

Please make my answer the braniest

Answer 2

${\huge{\underline{\underline{\rm{Question:-}}}}}$

Q) If the real roots exists of the following Quadratic Equation 2x² - 5x - 3 = 0 Then find the roots of the Equation .

${\huge{\underline{\underline{\rm{Answer\checkmark}}}}}$

$\bf{Given}\begin{cases}\sf{Equation→}\bf{2x^2-5x-3=0} \\ \rm{The \; Equation\; has \; real \; roots } \end{cases}$

★ To Find :

• The roots of the Equation .

★ Solution :

We can Solve it using two methods :

${\underline{\textit{\textbf{1st\;Middle\;Term\;Splitting}}}}$

$\mapsto \rm \: 2 {x}^{2} - 5x - 3 = 0$

$\mapsto \rm \: 2 {x}^{2} - 6x + x - 3 = 0$

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

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

Either :

$\rightarrow \rm \: 2x + 1 = 0$

$\rightarrow \rm \: 2x = - 1$

$\rightarrow \: {\underline{ \boxed{ \pink{ \rm{ x = \frac{ - 1}{2} }}}}}$

Or :

$\rightarrow \rm \: x - 3 = 0$

$\rightarrow \underline{ \boxed{ \pink{ \rm{x = 3}}}}$

So ,

# Roots : $\bold{\dfrac{-1}{2}}$ and 3 .

_________________________

${\underline{\textit{\textbf{2nd\; Quadratic\;Formula}}}}$

In this equation ,

$\boxed{ \sf{2 {x}^{2} - 5x - 3 = 0}}$

Compare it with standard equation

$\boxed{ \sf{a {x}^{2} + bx + c = 0}}$

So ,

• a = 2
• b = -5
• c = -3

Using the Quadratic Formula :

$\boxed{ \sf{roots = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}}}$

$\mapsto \rm \: roots = \frac{ - ( - 5) \pm \: \sqrt{ {( - 5)}^{2} - 4(2)( - 3)} }{2 \times 2}$

$\mapsto \rm \: roots = \frac{5 \pm \: \sqrt{25 + 24} }{4}$

$\mapsto \rm \: roots = \frac{5 \pm \: \sqrt{49} }{4}$

$\mapsto \rm \: roots = \frac{5 \pm7}{4}$

So ,

$\rightarrow \rm \: first \: root = \frac{5 + 7}{4}$

$\rightarrow \rm \: first \: root = \frac{ \cancel{12}}{ \cancel4}$

$\rightarrow \underline{ \boxed{ \pink{ \rm {first \: root = 3}}}}$

And ,

$\rightarrow \rm \: second \:root= \frac{5 - 7}{2}$

$\rightarrow \rm \: second \:root= \frac{ - \cancel2}{ \cancel4}$

$\rightarrow \underline{ \boxed{ \pink{ \rm{second \: root = \frac{ - 1}{2} }}}}$

So ,

# Roots : $\bold{\dfrac{-1}{2}}$ and 3 .

So , By both methods , we get same answer .