Polynomials Class 10th Important Questions with Answers Mathematics
Updated on June 1, 2025 | By Learnzy Academy
Q1. If one zero of the polynomial (a² + 9)x² + 13x + 6a is the reciprocal of the other, find the value of a.
If one zero of the polynomial (a² + 9)x² + 13x + 6a is the reciprocal of the other, find the value of a.
Step 1: Use the relationship between the roots
Let the roots be α and β, and it's given that α = 1/β.
Product of roots of a quadratic: αβ = c/a = 6a / (a² + 9)
Since α = 1/β, we get: αβ = 1
Step 2: Solve the equation
6a / (a² + 9) = 1
⇒ 6a = a² + 9
⇒ a² - 6a + 9 = 0
Step 3: Factor the quadratic
a² - 6a + 9 = (a - 3)² = 0
Therefore, a = 3
Q2. Find the zeroes of the polynomial: p(x) = x² − 7x + 10 and verify the relation between zeroes and coefficients.
Q3. Find a quadratic polynomial whose zeroes are 5 and −3.
Q4. If the zeroes of the quadratic polynomial p(x) = ax² + bx + c are reciprocal of each other, prove that c = a.
Q5. If the product of zeroes of the polynomial p(x)=3x² + kx − 2 is 2/3 , find the value of k.
Q6. Find the value of "p" from the polynomial x² + 3x + p, if one of the zeroes of the polynomial is 2.
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
Put x = 2 in equation x2 + 3x + p
⇒ 22 + 3(2) + p = 0
⇒ 4 + 6 + p = 0
⇒ 10 + p = 0
⇒ p = -10
Hence P = -10 is the answer
Q7. Find the zeroes of the polynomial 4x^2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.
Q8. Find the quadratic polynomial if its zeroes are 0 and √5.
A quadratic polynomial with zeros α and β can be written as with k is a non zero constant :
k(x2 – (α + β)x + αβ)
Where α and β are the roots of the polynomial.
Here it is given that α = 0 and β = √5
So, the polynomial will be:
k(x2 – (0 + √5)x + 0(√5))
k(x2 – √5x)
Let k = 1 for simplicity the equation is x2 – √5x
Q9. α and β are zeroes of the quadratic polynomial x² – 6x + y. Find the value of ‘y’ if 3α + 2β = 20
Let, f(x) = x² – 6x + y
And it is given that
3α + 2β = 20———————(i)
As α and β are zeroes of the quadratic polynomial x² – 6x + y
So α + β = -b/a = -(-6)/1
=> 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
Substitute the values of α and β in equation (iii) to get the value of y, such as;
y = αβ = (8)(-2) = -16
Hence value of y is -16