find the quadratic polynomial for the zeroes alpha and bheta given in each case root 3,-root3
Answers
Step-by-step explanation:
Answer:
(i)x^2-x-2x
2
−x−2
(ii)x^2-3x
2
−3
(iii)4x^2+3x-14x
2
+3x−1
(iv)4x^2-8x+34x
2
−8x+3
Step-by-step explanation:
(i)We are given that the zeroes of quadratic polynomial
\alpha=2,\bet=-1α=2,\bet=−1
We have to find the quadratic polynomial
We know that general quadratic polynomial
x^2-(\alpha+\beta)x+ \alpha\times \betax
2
−(α+β)x+α×β
Substituting the values of zeroes in general polynomial
Then we get x^2-(2-1)x+2\times(-1)x
2
−(2−1)x+2×(−1)
x^2-x-2x
2
−x−2
Hence, the quadratic polynomial is given by
x^2-x-2x
2
−x−2
(ii)We are given that zeroes of quadratic polynomial
\alpha=\sqrt3,\beta=-\sqrt3α=
3
,β=−
3
Substituting the values of zeroes in the general quadratic polynomial
Then we have x^2-(\sqrt3-\sqrt3)x-\sqrt3\times \sqrt3x
2
−(
3
−
3
)x−
3
×
3
x^2-3x
2
−3
Therefore, the quadratic polynomial
x^2-3x
2
−3
(iii) We are given that two zeroes of quadratic polynomial
\alpha=\frac{1}{4},\beta=-1α=
4
1
,β=−1
Substituting the values of zeroes in the general quadratic polynomial
Then we have x^2-(\frac{1}{4}-1)x-\frac{1}{4}x
2
−(
4
1
−1)x−
4
1
x^2-\frac{1-4}{4}x-\frac{1}{4}x
2
−
4
1−4
x−
4
1
Substituting the equation equal to zero
x^2+\frac{3}{4}x-\frac{1}{4}=0x
2
+
4
3
x−
4
1
=0
\frac{4x^2+3x-1}{4}=0
4
4x
2
+3x−1
=0
Hence, the quadratic polynomial
4x^2+3x-14x
2
+3x−1
(iv) We are given that two zeroes of quadratic polynomial
\alpha=\frac{1}{2},\bet=\frac{3}{2}α=
2
1
,\bet=
2
3
Substituting the values of zeroes in the general quadratic polynomial
Then we have
x^2-(\frac{1}{2}+\frac{3}{2})x+\frac{1}{2}\times\frac{3}{2}x
2
−(
2
1
+
2
3
)x+
2
1
×
2
3
x^2-\frac{1+3}{2}x+\frac{3}{4}x
2
−
2
1+3
x+
4
3
Substituting equation equal to zero
x^2-2x+\frac{3}{4}=0x
2
−2x+
4
3
=0
\frac{4x^2-8x+3}{4}=0
4
4x
2
−8x+3
=0
4x^2-8x+3=04x
2
−8x+3=0
Hence, the quadratic polynomial
4x^2-8x+34x
2
−8x+3