How do you Factor this equation
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Question: Given that BE bisects ∠CEA, which statements must be true? Select THREE options.
(See attachment below for the figure)
m∠CEA = 90°
m∠CEF = m∠CEA + m∠BEF
m∠CEB = 2(m∠CEA)
∠CEF is a straight angle.
∠AEF is a right angle.
Answer:
m∠CEA = 90°
∠CEF is a straight angle.
∠AEF is a right angle
Step-by-step explanation:
Line AE is perpendicular to line CF, which is a straight line. This creates two right angles, <CEA and <AEF.
Angle on a straight line = 180°. Therefore, m<CEA + m<AEF = m<CEF. Each right angle measures 90°.
Thus, the three statements that must be TRUE are:
m∠CEA = 90°
∠CEF is a straight angle.
∠AEF is a right angle
Answer:
1) Real and same.
2) Real and distinct
3) Real and distinct
4) Real and distinct
5) Real and distinct
6) Real and distinct
7) Real and distinct
8) Real and distinct
Step-by-step explanation:
If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.
If D > 0, then there will be two different real roots.
If D = 0, then two same and real roots.
If D < 0, then two distinct but imaginary roots.
Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.
2) 5x² - x = 4x² + 2
⇒ x² - x - 2 = 0
It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.
3)
⇒ 5 = x² - 4x
⇒ x² - 4x - 5 = 0.
So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.
4) x(x - 5) = 4(5 - x)
⇒ x² - x - 20 = 0
Hence, D = (-1)² - 4 × 1 × (-20) = 81
So, the roots will be real and distinct.
5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45
So, the roots will be real and distinct.
6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.
Hence, the roots will be real and distinct.
7) 5x² - 6 = 13x
⇒ 5x² - 13x - 6 = 0
So, D = (-13)² - 4 × 5 × (-6) = 289
So, the roots will be real and distinct.
8) x² - x = 3(x + 7)
⇒ x² - 4x - 21 = 0
It has D = (-4)² - 4 × 1 × (-21) = 100
So, the roots will be real and distinct. (Answer)