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Byju's Answer

Standard VIII

Mathematics

Finding Square Root of a Number

If t > 0 an...

Question

If $t > 0$ and $t^{2} - 4 = 0$ , what is the value of $t$ ?

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Solution

The correct option is A $2$
Given, $t^{2} - 4 = 0$
$\Rightarrow (t - 2) (t + 2) = 0$
$\Rightarrow t - 2 = 0$
$\Rightarrow t = 2$
Or $t + 2 = 0$
$\Rightarrow t = - 2$
But $t > 0$ , therefore $t = 2$
Option A is correct.

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Similar questions

Q. If

t^{2} + t + 1 = 0,

then the value of the expression

(t + \frac{1}{t}) + (t^{2} + \frac{1}{t^{2}}) + \dots + (t^{27} + \frac{1}{t^{^{27}}})

equals

Q. If

t^{2} + t + 1 = 0

,then value of

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}

is equal to

Q. If

t^{2} + t + 1

= 0 then prove that the value of

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . {(t^{27} + \frac{1}{t^{27}})}^{2} = 54

Q. If

x = \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}

and

y = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}},

then the value of

\frac{d^{2} y}{d x^{2}}

t = 0

is given by

Find the values of 't' in the equation $x^{2}$ - 2tx + $t^{2}$ - 1 = 0 such that exactly one root lies in between the numbers 2 and 4, and no root of the equation is either 2 (or) 4.

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If t>0 and t2−4=0, what is the value of t?

The correct option is A 2Given, t2−4=0⇒(t−2)(t+2)=0⇒t−2=0⇒t=2 Or t+2=0⇒t=−2But t>0, therefore t=2Option A is correct.

If $t > 0$ and $t^{2} - 4 = 0$ , what is the value of $t$ ?

The correct option is A $2$
Given, $t^{2} - 4 = 0$
$\Rightarrow (t - 2) (t + 2) = 0$
$\Rightarrow t - 2 = 0$
$\Rightarrow t = 2$
Or $t + 2 = 0$
$\Rightarrow t = - 2$
But $t > 0$ , therefore $t = 2$
Option A is correct.