1) Find the zeroes of the following quadratic polynomial and verify the relationship
between the zeroes and the co-efficient of 2x2 -3 + 5x.
2). If a and B are the zeroes of the polynomialoy
7y + 2, find a quadratic polynomial whose zeroes are 1/a and 1/b.
3. If the product of zeroes of the polynamial ax-6x-6 is 4, find the value of 'a'.
4.Find the quadratic polynomial whose zeroes are -2/3 and -√3/2
5. Find the zeroes of the polynomial x² + 3x-2 and verify the relationship between the
coefficients and relationship and the zeroes of the polynomial.
6. Find the value of m if the one zero of the polynomial (m +4]x2 + 6
reciprocal of the other.
7. Divide : 3x2-x2-3x + 5 by x-1-x2, and verify the division algorithem.
Answers
Answer:
i) x
2
−2x−8
Factorize the equation, we get (x+2)(x−4)
So, the value of x
2
−2x−8 is zero when x+2=0,x−4=0, i.e., when x=−2 or x=4.
Therefore, the zeros of x
2
−2x−8 are -2 and 4.
Now,
⇒Sum of zeroes =−2+4=2=−
1
2
=−
Coefficient of x
2
Coefficient of x
⇒Product of zeros =(−2)×(4)=−8 =
1
−8
=
Coefficient of x
2
Constant term
(ii) 4s
2
−4s+1
Factorize the equation, we get(2s−1)(2s−1)
So, the value of 4s
2
−4s+1 is zero when 2s−1=0,2s−1=0, i.e., when s=
2
1
or s=
2
1
.
Therefore, the zeros of 4s
2
−4s+1 are
2
1
and
2
1
.
Now,
⇒Sum of zeroes =
2
1
+
2
1
=1=−
4
−4
=−
Coefficient of s
2
Coefficient of s
⇒Product of zeros =
2
1
×
2
1
=
4
1
=
4
1
=
Coefficient of s
2
Constant term
(iii) 6x
2
−3−7x
Factorize the equation, we get (3x+1)(2x−3)
So, the value of 6x
2
−3−7x is zero when 3x+1=0,2x−3=0, i.e., when x=−
3
1
or x=
2
3
.
Therefore, the zeros of 6x
2
−3−7x are −
3
1
and
2
3
.
Now,
⇒Sum of zeroes = −
3
1
+
2
3
=
6
7
=−
6
−7
=−
Coefficient of x
2
Coefficient of x
⇒Product of zeros = −
3
1
×
2
3
=−
2
1
=
6
−3
=
2
−1
=
Coefficient of x
2
Constant term
(iv) 4u
2
+8u
Factorize the equation, we get 4u(u+2)
So, the value of 4u
2
+8u is zero when 4u=0,u+2=0, i.e., when u=0 or u=−2.
Therefore, the zeros of 4u
2
+8u are 0 and −2.
Now,
⇒Sum of zeroes = 0−2=−2=−
4
8
=−2=−
Coefficient of u
2
Coefficient of u
⇒Product of zeros = −0x−2=0=
4
0
=0=
Coefficient of u
2
Constant term
(v) t
2
−15
Factorize the equation, we get t=±
15
So, the value of t
2
−15 is zero when t+
15
=0,t−
15
=0, i.e., when t=
15
or t=−
15
.
Therefore, the zeros of t
2
−15 are ±
15
.
Now,
⇒Sum of zeroes =
15
−
15
=0=−
1
0
=0=−
Coefficient of t
2
Coefficient of t
⇒Product of zeros =
15
×
−15
=−15=
1
−15
=
Coefficient of t
2
Constant term
(vi) 3x
2
−x−4
Factorize the equation, we get(x+1)(3x−4)
So, the value of 3x
2
−x−4 is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x =
3
4
.
Therefore, the zeros of 3x
2
−x−4 are -1 and
3
4
.
Now,
⇒Sum of zeroes = −1+
3
4
=
3
1
=−
3
−1
=−
Coefficient of x
2
Coefficient of x
⇒Product of zeros = −1×
3
4
=−
3
4
=
3
−4
=
Coefficient of x
2
Constant term