Question

Please solve Q3 (2) and Q1 (5)

Answer 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

Answer 2

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

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