give me some extra and important questions of polynomial for class 10th you can also provide me app for extra solution and
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Answer:
1. Find the value of “p” from the polynomial x2 + 3x + p, if one of the zeroes of the polynomial is 2.
Solution:
As 2 is the zero of the polynomial.
We know that if α is a zero of the polynomial p(x), then p(α) = 0
Substituting x = 2 in x2 + 3x + p,
⇒ 22 + 3(2) + p = 0
⇒ 4 + 6 + p = 0
⇒ 10 + p = 0
⇒ p = -10
2. Does the polynomial a4 + 4a2 + 5 have real zeroes?
Solution:
In the aforementioned polynomial, let a2 = x.
Now, the polynomial becomes,
x2 + 4x + 5
Comparing with ax2 + bx + c,
Here, b2 – 4ac = 42 – 4(1)(5) = 16 – 20 = -4
So, D = b2 – 4ac < 0
As the discriminant (D) is negative, the given polynomial does not have real roots or zeroes.
3. Compute the zeroes of the polynomial 4x2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.
Solution:
Let the given polynomial be p(x) = 4x2 – 4x – 8
To find the zeroes, take p(x) = 0
Now, factorise the equation 4x2 – 4x – 8 = 0
4x2 – 4x – 8 = 0
4(x2 – x – 2) = 0
x2 – x – 2 = 0
x2 – 2x + x – 2 = 0
x(x – 2) + 1(x – 2) = 0
(x – 2)(x + 1) = 0
x = 2, x = -1
So, the roots of 4x2 – 4x – 8 are -1 and 2.
Relation between the sum of zeroes and coefficients:
-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x2)
Relation between the product of zeroes and coefficients:
(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x2)
4. Find the quadratic polynomial if its zeroes are 0, √5.
Solution:
A quadratic polynomial can be written using the sum and product of its zeroes is:
x2 – (α + β)x + αβ
Where α and β are the roots of the polynomial.
Here, α = 0 and β = √5
So, the polynomial will be:
x2 – (0 + √5)x + 0(√5)
⇒ x2 – √5x
5. Find the value of “x” in the polynomial 2a2 + 2xa + 5a + 10 if (a + x) is one of its factors.
Solution:
Say, f(a) = 2a2 + 2xa + 5a + 10
Since, (a + x) is a factor of 2a2 + 2xa + 5a + 10, f(-x) = 0
So, f(-x) = 2x2 – 2x2 – 5x + 10 = 0
Or, -5x + 10 = 0
Thus, x = 2
6. How many zeros does the polynomial (x – 3)2 – 4 can have? Also, find its zeroes.
Solution:
Given equation is (x – 3)2 – 4
Now, expand this equation.
=> x2 + 9 – 6x – 4
= x2 – 6x + 5
As the equation has a degree of 2, the number of zeroes will be 2.
Now, solve x2 – 6x + 5 = 0 to get the roots.
So, x2 – x – 5x + 5 = 0
=> x(x – 1) -5(x – 1) = 0
=> (x – 1)(x – 5) = 0
x = 1, x = 5
So, the roots are 1 and 5.
7. α and β are zeroes of the quadratic polynomial x2 – 6x + y. Find the value of ‘y’ if 3α + 2β = 20.
Solution:
Let, f(x) = x² – 6x + y
From the given,
3α + 2β = 20———————(i)
From f(x),
α + β = 6———————(ii)
And,
αβ = y———————(iii)
Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),
=> α = 20 – 12 = 8
Now, substitute this value in equation (ii),
=> β = 6 – 8 = -2
put the value of α and β in equation (iii) to get the value of y, such as;
y = αβ = (8)(-2) = -16
8. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, then find the value of a and b.
Solution:
Let the given polynomial be:
p(x) = x3 – 3x2 + x + 1
Given,
The zeroes of the p(x) are a – b, a, and a + b.
Now, compare the given polynomial equation with general expression.
px3 + qx2 + rx + s = x3 – 3x2 + x + 1
Here, p = 1, q = -3, r = 1 and s = 1
For sum of zeroes:
Sum of zeroes will be = a – b + a + a + b
-q/p = 3a
Substitute the values q and p.
-(-3)/1 = 3a
Or, a = 1
So, the zeroes are 1 – b, 1, 1 + b.
For the product of zeroes:
Product of zeroes = 1(1 – b)(1 + b)
-s/p = 1 – 2
=> -1/1 = 1 – 2
Or, 2 = 1 + 1 =2
So, b = √2
Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 3 − 32 + + 1.
9. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4, -1
(ii) 1,1
(iii) 4, 1
Solution:
(i) From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β
Product of zeroes = αβ
Given,
Sum of zeroes = 1/4
Product of zeroes = -1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ
x2 – (1/4)x + (-1)
4x2 – x – 4
Thus, 4x2 – x – 4 is the required quadratic polynomial.
(ii) Given,
Sum of zeroes = 1 = α + β
Product of zeroes = 1 = αβ
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ
x2 – x + 1
Thus, x2 – x + 1 is the quadratic polynomial.
(iii) Given,
Sum of zeroes, α + β = 4
Product of zeroes, αβ = 1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 – (α + β)x + αβ
x2 – 4x + 1
Thus, x2 – 4x +1 is the quadratic polynomial.