MATH SOLVE

2 months ago

Q:
# Find the nature of the root1)x²+6x+9=02)5x²-x=4x²+23)2/x+3/x=x-44)x(x-5)=4(5-x)5)x²+7x+1=06)2x²+9x+3=07)5x²-6=13x8)x²-x=3(x+7)

Accepted Solution

A:

Answer:1) Real and same.2) Real and distinct3) Real and distinct4) Real and distinct5) Real and distinct6) Real and distinct7) Real and distinct8) Real and distinctStep-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) [tex]\frac{2}{x} + \frac{3}{x} = x - 4[/tex]
⇒ 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)