1)(x + a) is a factor of f(x) if_________
2)if (x+1) is a factor of ax? +bx+c then the relation between the coefficients_________
3)If x + y is a factor of x^n+ y^n (n€N) then 'n' is_______
4)If a ¥ b, condition for ax^2+ ax + b and x² + bx + a to have a common root is________
5) The product of zeros of 6x² = 1 is_______
6)The graph of x = y lies in
quadrants_______
7)The graph of y = mx? is a________
8)The condition for (x+1) to be a factor of ax' + bx? + cx+d is_________
9) The number of terms in a polynomial whose degree is 11 is.______
10)The degree of the polynomial (x+y) is________
11)The condition for one root of ax? + bx + c = 0 is reciprocal of the other is___________
.
12)________is a factor of f(x) if the sum of the coefficients of even powers of 'X' is equal to the sum of the coefficient of
odd numbers of 'X'
13) For a polynomial if the sum of the coefficients of even powers of 'X' is equal to the sum of the coefficients of odd
numbers of x then______
is a factor of f(x)
Answers
Answer:
The value of \bold{\sqrt{a}+\frac{1}{\sqrt{a}}}
a
+
a
1
is 4
Given:
a=7-4 \sqrt{3}a=7−4
3
To find:
The value of \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
Solution:
The given value of a is 7-4 \sqrt{3}7−4
3
And the value of \frac{1}{a} \text { is } 7+4 \sqrt{3}
a
1
is 7+4
3
Then,
\begin{gathered}\begin{array}{c}{a+\frac{1}{a}=7-4 \sqrt{3}+7+4 \sqrt{3}} \\ {a+\frac{1}{a}=14}\end{array}\end{gathered}
a+
a
1
=7−4
3
+7+4
3
a+
a
1
=14
Now to find the value of \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
,
By using the formula of (a+b)^{2}=a^{2}+b^{2}+2 a b(a+b)
2
=a
2
+b
2
+2ab
Then, applying the (a+b)^{2}(a+b)
2
formula to \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
We can get,
\begin{gathered}\begin{array}{c}{\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=(\sqrt{a})^{2}+\left(\frac{1}{\sqrt{a}}\right)^{2}+\left(2 \times \sqrt{a} \times \frac{1}{\sqrt{a}}\right)} \\ {\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=a+\frac{1}{a}+2}\end{array}\end{gathered}
(
a
+
a
1
)
2
=(
a
)
2
+(
a
1
)
2
+(2×
a
×
a
1
)
(
a
+
a
1
)
2
=a+
a
1
+2
Already we know the value of a+\frac{1}{a}a+
a
1
is 14
\begin{gathered}\begin{array}{c}{\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=14+2} \\ {\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=16}\end{array}\end{gathered}
(
a
+
a
1
)
2
=14+2
(
a
+
a
1
)
2
=16
By taking square root for the value of \left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}(
a
+
a
1
)
2
, we can get the value of \sqrt{a}+\frac{1}{\sqrt{a}}
a
+
a
1
\begin{gathered}\begin{array}{l}{\sqrt{a}+\frac{1}{\sqrt{a}}=\sqrt{16}} \\ {\sqrt{a}+\frac{1}{\sqrt{a}}=4}\end{array}\end{gathered}
a
+
a
1
=
16
a
+
a
1
=4
Then the value of \bold{\sqrt{a}+\frac{1}{\sqrt{a}}}
a
+
a
1
is 4, if the value of a is 7-4 \sqrt{3}7−4
3
.