The values of x in the equation
7(x+2p)^2 + 5p^2 = 35xp + 117p^2 are
a) (4p, -3p) b) (4p, 3p)
c) (-4p, 3p)
d) (-4p. -3p)
Answers
✪Answer :-
Option - A
✪Given to find the value of x in the equation :-
7(x + 2p)² + 5p² = 35xp + 117p²
✪Solution:-
Firstly lets simplify the L.H.S by using algebraic identity (a+b)² = a² + 2ab + b²
7(x)² + (2p)² + 2(x)(2p) + 5p² = 35xp + 117p²
7[x² + 4p² + 4xp ] + 5p² = 35xp + 117p²
7x² + 28p² + 28xp + 5p² = 35xp + 117p²
✪Tranposing R.H.S equation to L.H.S
7x² + 28p² + 28xp +5p² -35xp -117p² = 0
✪Keeping like terms together
7x² +28p² + 5p²- 117p² + 28xp - 35xp = 0
7x² - 84p² - 7xp = 0
✪Take common ' 7'
7[x² - 12p² - xp] = 0
x² - 12p² -xp = 0/7
x² - 12p² -x p = 0
x² -xp - 12p² = 0
Since , we got the Quadratic equation Lets find the roots of the Quadratic equation
x² - xp - 12p² = 0
✪Splitting the middle term
x² + 3xp -4xp - 12p² = 0
x(x + 3p) -4p(x +3p ) = 0
(x + 3p) (x -4p ) = 0
✪Finding the roots :-
x + 3p = 0
x = -3p
x - 4p = 0
x = 4p
So, the value of x are -3p , 4p [A]
✪Know more about Quadratic equation:-
✪The Quadratic equation is the equation having with degree 2
✪ The Quadratic equation has 2 roots
✪ The general form of Quadratic equation is ax² + bx + c =0
✪ We can find the two roots by different methods like factorisation method, Formula method , complete squaring etc
✪ We can find the nature of roots i.e [Complex, distinct, equal , real etc] by discriminant of the Quadratic equation
✪ Discriminant of the Quadratic equation is b² - 4ac
Answer:
Option (A) is the correct answer.
HOPE IT'S HELP YOU.