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If 5 is a zero of the quadratic polynomial, x^{2}  kx  15 then the value of k is:
p(x) = x^{2} kx  15
Given: p(5) = 0
⇒ (5)^{2} k(5)  15 = 0
⇒ 25  5k  15 = 0
⇒ 5k = 10
⇒ k = 10/5 = 2
Thus, Value of k is 2
If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as:
Let p(x) = ax + b
p(k) = ak + b = 0
∴ k is zero of p(x)
Given, p(x) = 2x+5
For zero of the polynomial, put p(x) = 0
⇒ 2x + 5 = 0
⇒ x = 5/2
Thus, zero of the polynomial p(x) is 5/2.
If one of the zeroes of the quadratic polynomial (k  1)x^{2} + kx + 1 is 3, then the value of k is:
p(x) = (k  1)x^{2} + kx +1
One zero of polynomial is  3 i.e. p(3) = 0
⇒ (k  1) (3)^{2} + k(3) + 1 = 0
⇒ 9k  9  3k + 1 = 0
⇒ 6k  8 = 0
⇒ k = 8/6 = 4/3
If one of the zeros of a quadratic polynomial of the form x^{2} + ax + b is the negative of the other, then it
Let p(x) = x^{2} + ax + b.
Put a = 0 then, p(x) = x^{2} + b = 0
⇒ x^{2 }= b
⇒ x = ±√b [∴ b < 0]
Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other,
then it has no linear term i.e., a = 0 and the constant term is negative i.e., b< 0.
Alternate Method
Let f(x) = x^{2} + ax+ b
Given condition the zeroes are α and – α.
Sum of the zeroes (a) = α  α = 0
Product of zeroes (b) = α .( α) =  α^{2}
Hence, it has no linear term and the constant term is negative.
If the zeroes of the quadratic polynomial x^{2} + (a + 1) x + b are 2 and 3, then
p(x) = x^{2} + (a + 1)x + b
Given: zeros of polynomial is 2, 3
For x = 2
(2)^{2} + (a + 1)^{2} + b = 0
⇒ 4 + 2a + 2 + b = 0
⇒ 2a + b = 6 ...(i)
For x = 3
(3)^{2} + (a + 1) (3) + b = 0
⇒ 9  3a  3 + b = 0
⇒ 3a + b = 6 ..(ii)
Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = 6
Alternative method:
p(x) = x^{2} + (a + 1)x + b
Given: zeros of polynomial is 2, 3
Sum of zeros: (a+1) = 23
⇒ a = 0
Product of zeros: b = 2.(3) = 6
The number of polynomials having zeros as  2 and 5 is:
Let p (x) = ax^{2} + bx + c be the required polynomial whose zeroes are 2 and 5.
∴ Sum of zeroes = b/a
⇒  2 + 5 = 3 = (3)/1 ...(i)
and product of zeroes = c/a
⇒ 2 x 5 = 10/1 ...(ii)
From Eqs. (i) and (ii)
a = 1, b = 3 and c = 10
∴ p(x) = ax^{2} + bx + c = 1.x^{2}  3x  10
= x^{2}  3x  10
But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.
∴ p(x) = kx^{2}  3kx  10k [where, k is a real number]
⇒ p(x) = x^{2}/k  3x/k 10/k [where, k is a nonzero real number]
Hence, the required number of polynomials are infinite i.e., more than 3.
Which of the following is NOT the graph of a quadratic polynomial ?
For any quadratic polynomial ax^{2} + bx + c, a≠0, graph for the corresponding equation:
Thus, option A is correct.
If one root of the polynomial p(y) = 5y^{2} + 13y + m is reciprocal of other, then the value of m is
Let the roots be α and 1/α.
Product of roots = α(1/α) = 1
⇒ m/5 = 1
⇒ m = 5
Sum of zeroes = b/a
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