Class X (CBSE 2019) Mathematics Abroad (Set-2)

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Question 1

For what values of k does the quadratic equation <math>4x^2 - 12x - k = 0</math> have no real roots?

SOLUTION:

We have been given the quadratic equation:

<math>4x^2 - 12x - k = 0</math>

To have no real roots means discriminant should be less than zero.

<math>D=b^2-4ac</math>

<math>b^2 - 4ac < 0</math>

Plugging the values in the formula of discriminant

<math>(-12)^2 - 4(4)(-k) < 0</math>

<math>144 + 16k < 0</math>

<math>k < -9</math>

Therefore, for k<-9 the quadratic equation will have no real roots.

Question 2

Find the distance between the points <math>(a, b)</math> and <math>(-a, -b)</math>.

SOLUTION:

Using distance formula:

<math display="block">d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}</math>

Here, <math>x_1 = a</math>, <math>y_1 = b</math>, <math>x_2 = -a</math> and <math>y_2 = -b</math>

On substituting the values in the formula we get <math>\sqrt{(-a-a)^2+(-b-b)^2}</math> <math>=\sqrt{(-2a)^2+(-2b)^2}</math> <math>=\sqrt{4a^2+4b^2}</math> <math>=2\sqrt{a^2+b^2}</math>

Therefore, the distance between <math>(a, b)</math> and <math>(-a, -b)</math> is <math>2\sqrt{(a)^2+(b)^2}</math>

Question 3

Find a rational number between <math>\sqrt{2}</math> and <math>\sqrt{7}</math>. OR Write the number of zeroes in the end of a number whose prime factorization is <math>2^2 \times 5^3 \times 3^2 \times 17</math>.

SOLUTION:

We know

<math>\sqrt{2} = 1.414</math>

<math>\sqrt{7} = 1.732</math>

So, rational number between <math>\sqrt{2}</math> and <math>\sqrt{7}</math> will be 1.5 = <math>\frac{3}{2}</math>.

OR Given prime factorisation is <math>2^2 \times 5^3 \times 3^2 \times 17</math>.

A number will have zero at the end when we have <math>2 \times 5</math>.

In <math>2^2 \times 5^3 \times 3^2 \times 17</math> we will have 2 zeroes as <math>\left(2^2 \times 5^2\right) \times 5 \times 3^2 \times 17</math> .