solve the quadratic equation 2x square-5x-3=0
Answers
Answer:
2x
2
−5x+3=0
⇒x
2
−
2
5
x+
2
3
=0 (Dividing throughough by 2)
⇒x
2
−
2
5
x=−
2
3
(Shifting the constant term on RHS)
⇒x
2
−2(
4
5
)x+(
4
5
)
2
=(
4
5
)
2
−
2
3
(Adding (
2
1
Coeff.ofx)
2
onboth sides)
⇒(x−
4
5
)
2
=
16
25
−
2
3
⇒(x−
4
5
)
2
=
16
1
⇒x−
4
5
=±
4
1
⇒x=
4
5
±
4
1
⇒x=
4
5
+
4
1
=
4
6
orx=
4
5
−
4
1
=
4
4
⇒x=
2
3
orx=1
Here the roots of the equation are
2
3
and 1
Question:
solve the quadratic equation 2x square-5x-3=0
Answer:
Solution :-
Here, we have
⇒ 2x² - 5x + 3 = 0
Dividing equation by 2, we get
⇒ x² - 5/2x + 3/2 = 0
Shifting the constant term on RHS
⇒ x² - 5/2x = - 3/2
Adding (1/2 coefficient of x)² on both sides
⇒ (x - 5/4)² = 25/16 - 3/2
⇒ (x - 5/2)² = 1/16
⇒ x - 5/4 = ± 1/4 or x = 5/4 - 1/4 = 4/4
⇒ x = 3/2, 1
Hence, the roots of the equation 2x² - 5x + 3 = 0 are 3/2 and 1.