The solutions of a quadratic function are the values of x for which f(x) = 0. Solutions of functions are also called roots or zeros of the function.
On a graph, the solution of the function is the x-intercept(s).
Problem 1 :
Find zeroes of the quadratic polynomial
x^{2} + x – 2 = 0
a. 4, -5 b. 2, -4 c. 2, -1 d.1, -2
Solution :
Given, x^{2} + x – 2 = 0
x^{2} – x + 2x – 2 = 0
x(x – 1) + 2(x – 1) = 0
(x + 2) (x – 1) = 0
x + 2 = 0 and x – 1 = 0
x = -2 x = 1
So, option (d) is correct.
Problem 2 :
Write a quadratic equation with the given roots. Write the equation in the form ax^{2} + bx + c = 0, where a, b, and c are integers.
-5 and -1
Solution :
The given values are solution. That is values of x.
x = -5 and x = -1
The factors are x + 5 and x + 1
Multiplying the factors, we get
f(x) = (x + 5)(x + 1)
f(x) = x^{2} + 5x + x + 5
f(x) = x^{2} + 6x + 5
So, the required polynomial which are having two factors is
f(x) = x^{2} + 6x + 5
Problem 3 :
Write a quadratic polynomial, sum of whose zeroes is 2√3 and product is 5.
Solution :
Sum of zeroes = 2√3
Product of zeroes = 5
x^{2} - (α + β)x + αβ = 0
x^{2} - (Sum of zeroes)x + Product of zeroes = 0
x^{2} - 2√3x + 5 = 0
Problem 4 :
For what value of k, (–4) is a zero of the polynomial
x^{2} – x – (2k + 2)?
Solution :
Let f(x) = x^{2} – x – (2k + 2)
Since -4 is zero of the polynomial,
f(-4) = 0
f(4) = 4^{2} – 4 – (2k + 2) = 0
16 - 4 - 2k - 2 = 0
-2k + 10 = 0
-2k = -10
k = 5
So, the value of k is 5.
Problem 5 :
For what value of p, (–4) is a zero of the polynomial
x^{2} – 2x – (7p + 3)?
Solution :
Let f(x) = x^{2} – 2x – (7p + 3)
Since -4 is zero of the polynomial,
f(-4) = 0
f(4) = (-4)^{2} – 2(-4) – (7p + 3) = 0
16 + 8 - 7p - 3 = 0
21 - 7p = 0
7p = 21
p = 3
So, the value of p is 3.
Problem 6 :
If 1 is a zero of the polynomial p(x) = ax^{2} – 3(a – 1) x – 1, then find the value of a.
Solution :
Let p(x) = ax^{2} – 3(a – 1) x – 1
Since 1 is zero of the polynomial,
p(1) = a(1)^{2} – 3(a – 1)(1) – 1 = 0
a - 3a + 3 - 1 = 0
-2a + 2 = 0
-2a = -2
a = 1
So, the value of a is 1.
Problem 7 :
If (x + a) is a factor of 2x^{2} + 2ax + 5x + 10 find a.
Solution :
Let p(x) = 2x^{2} + 2ax + 5x + 10
If x + a = 0, then x = -a. p(-a) = 0
p(-a) = 2(-a)^{2} + 2a(-a) + 5(-a) + 10 = 0
2a^{2} - 2a^{2} - 5a + 10 = 0
-5a + 10 = 0
-5a = -10
a = 2
So, the value of a is 2.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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