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Question

The roots of the equation $y^2 - \sqrt{5} y - y + \sqrt{5} = 0$ are:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\sqrt{5};\ 1$

Solving for Roots of $y^2 - \sqrt{5} y - y + \sqrt{5} = 0$

To find the roots of the equation $y^2 - \sqrt{5} y - y + \sqrt{5} = 0$, we can use factoring by grouping.

First, group the terms of the equation:

$(y^2 - \sqrt{5} y) + (-y + \sqrt{5}) = 0$

Next, factor out the greatest common divisor from each group:

$y(y - \sqrt{5}) - 1(y - \sqrt{5}) = 0$

Notice that $(y - \sqrt{5})$ is a common binomial factor. Factor it out:

$(y - \sqrt{5})(y - 1) = 0$

For the product of two factors to equal zero, at least one of the factors must be zero. Set each factor equal to zero and solve for $y$:

  • Setting the first factor to zero: $y - \sqrt{5} = 0 \implies y = \sqrt{5}$
  • Setting the second factor to zero: $y - 1 = 0 \implies y = 1$

The roots of the equation are $\sqrt{5}$ and $1$.

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Important Questions from Quadratic Equation

  1. If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\)  is:

  2. If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:

  3. If x 2 – 3x + 1 = 0, then the value of  \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\)  is:

  4. If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) \(x \ne 0\) , then what is the value of  \((x^4+{1\over{x^2}})\over(x^2+1) \)  ?

  5. If x 2\(\frac{1}{x^2}\)  = 18, x > 0, then find the value of x \(\frac{1}{x^3}\) .

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