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Question
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
`2x^2-2sqrt2x+1=0`
Solution
We have been given, `2x^2-2sqrt2x+1=0`
Now we also know that for an equation ax2 + bx + c = 0, the discriminant is given by the following equation:
D = b2 - 4ac
Now, according to the equation given to us, we have,a = 2, `b=-2sqrt2` and c = 1.
Therefore, the discriminant is given as,
`D=(-2sqrt2)^2-4(2)(1)`
= 8 - 8
= 0
Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real and equal roots.
Now, the roots of an equation is given by the following equation,
`x=(-b+-sqrtD)/(2a)`
Therefore, the roots of the equation are given as follows,
`x=(-(-2sqrt2)+-sqrt0)/(2(2))`
`=(2sqrt2)/4`
`=1/sqrt2`
Therefore, the roots of the equation are real and equal and its value is `1/sqrt2`
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