Question

will be a perfect square -
b) For what value of 't',
${u}^{2}$– tu + 1/4
i) 4
ii)8
iii)1
iv)1/2​

Answer 1

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$\underline{\huge{\frak{\red{question}}}}$

For what value of 't',

$\sf{ {u}^{2} - tu + \frac{1}{4}}$

will be a perfect square,

Options

(i) 4

(ii) 8

(iii) 1 ✔✔

(iv) 1/2

$\underline{\huge{\frak{\red{solution}}}}$

If

$\sf \: {u}^{2} - tu + \frac{1}{4}$

will be a perfect square then,

it will have two equal and real roots

such, that

$\boxed{ \sf \: {b}^{2} - 4ac = 0}$

In the given quadratic polynomial in the 'u' variable

a = 1 ; b = -t ; c = 1 /4

so, putting values in

$\sf \: {b}^{2} - 4ac = 0 \\ \\ \sf \: {( - t)}^{2} - 4(1)( \frac{1}{4} ) = 0 \\ \\ \sf \: {t}^{2} - 1 = 0 \\ \\ \sf \: {t}^{2} = 1 \\ \\\boxed{ \sf \: t = 1}$

Hence the value of t will be 1 to make

u^2 - tu + 1/4 a perfect square.

$\underline{\huge{\red{\frak{ correct \: answer}}}}$

So,

The correct answer to this question will be 1.

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