Question

Determine the nature of the roots of each of the
following equation :
(i) x² - 7x+12=0 (ii) 9x² - 6x +1 = 0
(iii) 2x²-x+1=0​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\huge\red{\sf{ꜱᴏʟᴜᴛɪᴏɴ:}}$

☞ (i) x²-7x+12 = 0

• x²-7x+12 = 0
• x²-3x-4x+12 = 0
• x(x-3)-4(x-3) = 0
• (x-4)(x-3) = 0
• x = 4 (or) 3

→ Value of x is 3 (or) 4

________________________

☞ (ii) 9x²-6x+1 = 0

• 9x²-6x+1 = 0
• 9x²-3x-3x+1 = 0
• 3x(3x-1)-1(3x-1) = 0
• (3x-1)(3x-1) = 0
• x = 1/3 (or) 1/3

→ Value of x is 1/3, Here the roots are Real and Equal.

________________________

☞ (iii) 2x²-x+1 = 0

2x²-x+1 = 0 is in the form of ax²+bx+c = 0

• a = 2
• b = -1
• c = 1

ʙʏ ᴅʜᴀʀᴀᴄʜᴀʀʏᴀ ꜰᴏʀᴍᴜʟᴀ

$\boxed{\sf{☆\;x = \frac{-b±\sqrt{b²-4ac}}{2a}}}$

$→\;{\frac{-(-1)±\sqrt{(-1)²-4×2×1}}{2×2}}$

$→\;{\frac{1±\sqrt{1-8}}{4}}$

$→\;{\frac{1±\sqrt{7}}{4}}$

→ Value of x is ${\frac{1+\sqrt{7}}{4}}$ (or) ${\frac{1-\sqrt{7}}{4}}$

$\Large{✓}$

Hope It Helps You ✌️