Question

solve the quadratic equations by the formula method : 1p2-1p-3=0

formula
x = -b+-√b2-4ac÷2a​

Answer 1

EQUATION -:

$1 {p}^{2} + p - 3 = 0$

root formula -:

$\frac{ - b \frac{ + }{ - } \sqrt{d} }{2a}$

VALUE OF D =

$= > {b}^{2} - 4ac \\ \\ = > {1}^{2} - 4 \times ( - 3) \times 1 \\ \\ = > 13$

INSERT IN D VALUE IN FORMULA

(1)

$= > \frac{{( - 1)}^{2} + \sqrt{13} }{2 \times 1} \\ \\ = > \frac{1 + \sqrt{3} }{2}$

(2)

$= > \frac{ { - 1}^{2} - \sqrt{13} }{2 \times 1} \\ \\ = > \frac{1 - \sqrt{13} }{2}$

$\bf\underline{By-Saksham }$

$\sf{I\: Hope\: It\: Is\: Helpful\: For\: you!} \\ \\ </p><p></p><p>\huge\mathfrak\red{thank\:u}$

Answer 2

$\huge\mathfrak\red{Solution}$

Here the given equation is :

• 1p² - 1p - 3 = 0

We have,

a = 1

b = -1

c = - 3

• First we will find b² - 4ac

= b² + 4ac

= (-1)² - 4 * 1 * (-3)

= 1 + 12

= 13

• It means the value of b² - 4ac is √13

• By quadratic formula,

p = -b +- √b²-4ac / 2a

p = -(-1) +- √13 / 2 * 1

p = 1 +- √13 / 2

p = 1 + √13 / 2 ,

p = 1 - √13 / 2

Here there are two values first one we have to be with positive sign and second one with Negative sign.

• Note = In formula it is x but in question we take P because we have to find the value of P not of x that's why.

And I try my best to solve this and sorry if it is not helpful to you have a great future ahead ☺ ♥