Question

1. which equation has 1 real, rational root?

a. ( x^2 - 2x + 5 = 0 )
b. ( x^2 - 5x + 4 = 0 )
c. ( x^2 - 3x + 1 = 0 )
d. ( x^2 + 10x + 25 = 0 )

Explanation:

Step1: Recall discriminant formula

For quadratic equation \(ax^2 + bx + c = 0\), discriminant \(D = b^2 - 4ac\). If \(D = 0\), one real root; if \(D>0\), two real roots; if \(D<0\), no real roots. Also, root is rational if \(D\) is perfect square.

Step2: Analyze Option A (\(x^2 - 2x + 5 = 0\))

Here, \(a = 1\), \(b = -2\), \(c = 5\).
\(D = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0\). No real roots.

Step3: Analyze Option B (\(x^2 - 5x + 4 = 0\))

\(a = 1\), \(b = -5\), \(c = 4\).
\(D = (-5)^2 - 4(1)(4) = 25 - 16 = 9 > 0\). Two real roots (since \(D>0\)), not one.

Step4: Analyze Option C (\(x^2 - 3x + 1 = 0\))

\(a = 1\), \(b = -3\), \(c = 1\).
\(D = (-3)^2 - 4(1)(1) = 9 - 4 = 5\). \(5\) is not a perfect square, and \(D>0\) (two roots), not one.

Step5: Analyze Option D (\(x^2 + 10x + 25 = 0\))

\(a = 1\), \(b = 10\), \(c = 25\).
\(D = (10)^2 - 4(1)(25) = 100 - 100 = 0\). One real root (since \(D = 0\)), and root is \(\frac{-b}{2a}=\frac{-10}{2}=-5\) (rational).

Answer:

D. \(x^2 + 10x + 25 = 0\)