Question

7. Find the roots of the equation
$2x {}^{2} - x + \frac{1}{8} = 0.$

Standard:- 10

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Answer 1

Hello !!!

This is your answer.

Given :
${2x}^{2} - x + \frac{1}{8} = 0.$

I'll solve in an easy manner.
First of all, Multiply Whole Equation by 8, we get
16x² - 8x + 1 = 0

Now, We have to find two numbers a and b such that a+b = -8 and ab= 16.
By looking at the prime factorisation of 16 = 2×2×2×2.
We get a = 4 And b = 4.

So,
16x² - 8x + 1 = 0
=> 16x² - 4x - 4x + 1 = 0
=> 4x (4x - 1) - 1 (4x - 1) = 0
=> (4x - 1)(4x - 1)
Comparing with 0,
4x - 1 = 0
x = 1/4.
Hence, the roots of the given polynomial are $\frac{1}{4} \: and\: \frac{1}{4}$.

Hope You Got It

Answer 2

Hey!!!!!!

Difficult Level : Very Hard

Chances of being asked in Board : 50%

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Let's solve

=> 2x² - x + 1/8

Taking 1/8 common

=> 1/8(16x² - 8x + 1)

Now only considering 16x² - 8x + 1

=> 16x² - 4x - 4x + 1

=> 4x(4x - 1) - (4x - 1)

=> (4x - 1)²

Combining with 1/8

=> (4x - 1)²/8 = 0

=> (4x - 1) = 0

=> x = 1/4 or x = 1/4 <<<<<< Answer

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Hope this helps ✌️

Happy Monsoon