Answer:
a) (x + 5) (x - 5)
b) (x + 5i) (x - 5i)
c) (x + (5i/2)) (x - (5i/2))
d) (x-1)(x-1)
e) x +i√3 +1) (x -i√3+1)
Step-by-step explanation:
To solve this, we will need to factorize each quadratic function making it equal to zero first and then proceeding to find x
a) f(x) = x²-25
x²-25 = 0
⇒(x + 5) (x - 5)
b) f(x)=x²+25
x² + 25 = 0
x²= -25
x = √-25
x = √25i
x = ±5i
⇒(x + 5i) (x - 5i)
c) f(x)=4x²+25
4x²+25 = 0
4x²= -25
x² = -25/4
x = ±√(-25/4)
x = ±(√25i)/2
x = ±5i /2
⇒(x + (5i/2)) (x - (5i/2))
d) f(x)=x²-2x+1
x²-2x+1 = 0
⇒(x - 1)²
e) f(x)=x²-2x+4
x²-2x+4 = 0
x²-2x = -4
x²-2x +1 = -4 +1
x²-2x + 1 = -3
(x-1)² +3 = 0
(x-1)²= -3
x-1 = √-3
x = ±√3i +1
⇒(x +i√3 +1) (x -i√3+1)
Answer:
The Proof is below.
Step-by-step explanation:
Given:
To Prove:
Proof:
In ΔLPM and ΔNPM
……….{Given}
……….{Given}
……….{Reflexive Property}
ΔLPM ≅ ΔNPM ….{ By Side-Side-Side congruence test}
∴ ∠LMP ≅ ∠NMP ...{Corresponding parts of congruent triangles (c.p.c.t).}.....( 1 )
Now In ΔLMQ and ΔNMQ
……….{Given}
∠LMQ ≅ ∠NMQ ..........{From 1 above}
……….{Reflexive Property}
ΔLMQ ≅ ΔNMQ ....{ By Side-Angle-Side Congruence test}
∴ ...{Corresponding parts of congruent triangles (c.p.c.t).}.....Proved
Answer:
0.01
Step-by-step explanation:
