f –4 is a root of the equation x² + px – 4 = 0 and the quadratic equation x² + px + k = 0 has equal roots, find the value of k.
Question
f – 4 is a root of the equation
x² + px – 4 = 0
and the quadratic equation
x² + px + k = 0
has equal roots, find the value of k
.
Solution
Given that f –4 is a root of the equation x² + px – 4 = 0, we can substitute f –4 into the equation to find the value of p.
Substituting f –4 into the equation, we get: (f –4)² + p(f –4) – 4 = 0
Expanding and simplifying the equation, we have: f² – 8f + 16 + pf – 4p – 4 = 0 f² + (p – 8)f + 12 – 4p = 0
Since f –4 is a root of the equation, we know that when we substitute f = –4 into the equation, it should equal zero. So, we can substitute f = –4 into the equation and solve for p:
(-4)² + (p – 8)(-4) + 12 – 4p = 0 16 – 4p + 32 – 4p + 12 – 4p = 0 60 – 12p = 0 12p = 60 p = 5
Now that we have found the value of p, we can use it to find the value of k in the equation x² + px + k = 0.
Since the quadratic equation x² + px + k = 0 has equal roots, we know that the discriminant (b² – 4ac) is equal to zero. In this case, a = 1, b = p, and c = k.
Substituting these values into the discriminant formula, we have: p² – 4(1)(k) = 0 p² – 4k = 0
Substituting the value of p we found earlier, we have: 5² – 4k = 0 25 – 4k = 0 4k = 25 k = 25/4
Therefore, the value of k is 25/4.
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