Given:
G = \sqrt{\frac{p + 1}{2p - 1}}
Determine the value(s) of $p$ such that $G$ will be as follows:
2.1.1 Undefined
2.1.2 Equal to zero
2.2 Determine for which value(s) of $k$ the equation $x^2 - k + 4 = 5x$ will have real roots. - NSC Technical Mathematics - Question 2 - 2019 - Paper 1
Question 2
Given:
G = \sqrt{\frac{p + 1}{2p - 1}}
Determine the value(s) of $p$ such that $G$ will be as follows:
2.1.1 Undefined
2.1.2 Equal to zero
2.2 Determine ... show full transcript
Worked Solution & Example Answer:Given:
G = \sqrt{\frac{p + 1}{2p - 1}}
Determine the value(s) of $p$ such that $G$ will be as follows:
2.1.1 Undefined
2.1.2 Equal to zero
2.2 Determine for which value(s) of $k$ the equation $x^2 - k + 4 = 5x$ will have real roots. - NSC Technical Mathematics - Question 2 - 2019 - Paper 1
Step 1
2.1.1 Undefined
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Answer
To find when G is undefined, we look for values of p that make the denominator zero.
The denominator 2p−1 must equal zero:
2p−1=0
Solving for p gives:
2p=1p=21
Thus, G is undefined when p=21.
Step 2
2.1.2 Equal to zero
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Answer
To find when G equals zero, we need the numerator to be zero:
p+1=0
Solving for p gives:
p=−1
Thus, G is equal to zero when p=−1.
Step 3
2.2
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Answer
First, we rearrange the equation to standard form:
x2−5x+(4−k)=0
Next, we need to find the discriminant b2−4ac to determine when the equation has real roots. For the quadratic equation ax2+bx+c, we have:
a=1
b=−5
c=4−k
Calculating the discriminant:
(−5)2−4(1)(4−k)≥025−4(4−k)≥025−16+4k≥04k+9≥0
Thus, 4k≥−9 or:
k≥−49
Hence, the equation will have real roots when k≥−49.
