Question

2x^2-3x-5=0 by Completing method​

Answer 1

Answer:

hii.

Step-by-step explanation:

given equation is ...

2x^2-3x-5=0.

2x^2-5x+2x-5=0.

2x(X-1)+5(X-1)=0

(x-1)(2x +5)=0.

then .. x=1, -5/2

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Answer 2

TO SOLVE THE QUADRATIC EQUATION

by Completion of Square Method.

Step 01 :-

Divide coefficient of x^2 on both LHS and RHS.

$2{x}^{2}-3x-5 = 0 \\ \\ \\ \\ \implies {x}^2- \frac{3x}{2}-\frac{5}{2}$

Step 02 :-

We will be multiplying and dividing 2 with the coeficient of x

${x}^{2}- 2 \times \frac{3}{4} \times x= \frac{5 }{2}$

Step 03 :-

We will be adding the square of coefficient of x except the multiplied 2 both the sides.

${x}^{2}- 2 \times \frac{3}{4} \times x + {\frac{3}{4}}^{2}=\frac{5}{2}+{\frac{3}{4}}^{2}$

Step 04 :-

Now in the LHS we will be applying the reverse identity of (a - b)^2.

${( x - \frac{3}{4})}^{2} = \frac{5}{2}+{\frac{3}{4}}^{2}$

Step 05 :-

We will be square rooting both the side then simply solve it to get the roots of the equation.

${(x + \frac{3}{4}) }^{2} = \frac{9}{16}+ \frac{5}{2} \\ \\ \\ \\ \implies {(x + \frac{3}{4}) }^{2} = \frac{9+40}{16}\\ \\ \\ \\ \implies {(x + \frac{3}{4}) }^{2} = \frac{49}{4}\\ \\ \\ \\ \implies x + \frac{3}{4}=\pm \frac{7}{4} \\ \\ \\ \\ \implies x = \frac{-3 \pm7}{4}\\ \\ \\ \\ \implies x = \frac{4}{4}or \frac{-10}{4}\\ \\ \\ \\ \implies \boxed{x = 1 :\ or :\ x = \frac{-5}{4}}$

Hence we got two roots of this Quadratic equation.

DIRECT FORMULA :-

$x = \frac{-b \pm \sqrt{{b}^2-4ac}}{2a}$