Math, asked by user3679, 3 months ago

Please solve Q3 (2) and Q1 (5)

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Answers

Answered by awnish30
1

Answer:

put the value of x in every question

Step-by-step explanation:

5(1/2×1/2)+2k×(1/2)-3=0

5×1/4+k-3=0

5/4+k-3=0

5/4+k=0+3

5/4+k=3

k=3-5/4

k=7/4

vote please

Answered by mathdude500
2

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ques \: 1(5) \:  \:  \:   \\ {x}^{2}  + 2 \sqrt{x}  + 7 = 0 \: \\  is \: not \: a \: quadratic \: equation \:  \\ as \: x \: is \: not \: free \: from \: radicals.

ques \: 3(2) \\ since \: x =  \frac{1}{2}  \: is \: a \: solution \: of \:  {5x}^{2}  + 2kx - 3 = 0 \\ therefore \: 5 {( \frac{1}{2}) }^{2}  + 2k \times  \frac{1}{2}  - 3 = 0 \\  \frac{5}{4}  + k - 3 = 0 \\  =  >  \: k = 3 -  \frac{5}{4}  \\  =  >  \: k =  \frac{7}{4}

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