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Georgia [21]
3 years ago
15

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

Mathematics
1 answer:
andriy [413]3 years ago
4 0

Answer:

If k = 1 or k = -\frac{1}{2}, there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

The signal of \bigtriangleup determines how many real roots an equation has:

\bigtriangleup > 0: Two real and different solutions

\bigtriangleup = 0: One real solution

\bigtriangleup < 0: No real solutions

In this problem, we have the following second order polynomial:

(k+1)x^{2} + 4kx + 2 = 0.

This means that a = k+1; b = 4k; c = 2

It has one solution if

\bigtriangleup = 0

b^{2} - 4ac = 0

16k^{2} -8(k+1) = 0

16k^{2} - 8k - 8 = 0

We can simplify by 8

2k^{2} - k - 1 = 0

The solution is:

k = 1 or k = -\frac{1}{2}

So, if k = 1 or k = -\frac{1}{2}, there is only one solution to the given quadratic equation.

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How would i approximate each irrational square root number to the nearest 0.05? 34, 82, 45, 104, -71, and -19
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