Q14: Find the quadratic polynomial whose zeroes are 3.-2, the graph of which passes through (0,6)
Answers
Answer:
Answer: The quadratic polynomial is \mathbf{-x^{2}+x+6}−x
2
+x+6
Step-by-step explanation:
1. Let quadratic polynomial is
\mathbf{y= ax^{2}+bx+c}y=ax
2
+bx+c ...1)
2. Since quadratic polynomial have zeros 3 and -2. It means
At x=3 , y=0 ...2)
and also
At x= -2 , y=0 ...3)
3. It is also given that point (0,6) lie on curve. It means
At x=0 , y =6 ...4)
4. Now from equation 1) and equation 4), we get
\mathbf{y= ax^{2}+bx+c}y=ax
2
+bx+c
\mathbf{6= a(0)^{2}+b\times 0+c}6=a(0)
2
+b×0+c
\mathbf{6= 0+0+c}6=0+0+c
So
c=6
5. Now equation 1) can be written as
\mathbf{y= ax^{2}+bx+c}y=ax
2
+bx+c
\mathbf{y= ax^{2}+bx+6}y=ax
2
+bx+6 ...5)
6. Now from equation 2) and equation 5)
\mathbf{y= ax^{2}+bx+6}y=ax
2
+bx+6
\mathbf{0= a(3)^{2}+b\times 3+6}0=a(3)
2
+b×3+6
\mathbf{0= 9a+3b+6}0=9a+3b+6
\mathbf{9a+3b=-6}9a+3b=−6
This can be written as
\mathbf{3a+b=-2}3a+b=−2 ...6)
7. Now from equation 3) and equation 5)
\mathbf{y= ax^{2}+bx+6}y=ax
2
+bx+6
\mathbf{0= a(-2)^{2}+b\times (-2)+6}0=a(−2)
2
+b×(−2)+6
\mathbf{0= 4a-2b+6}0=4a−2b+6
\mathbf{4a-2b=-6}4a−2b=−6
This can be written as
\mathbf{2a-b=-3}2a−b=−3 ...7)
8. On adding equation 6) and equation 7)
\mathbf{5a=-5}5a=−5
\mathbf{a=-1}a=−1 ...8)
9. Now from equation 8) and equation 7)
\mathbf{2a-b=-3}2a−b=−3
\mathbf{2(-1)-b=-3}2(−1)−b=−3
\mathbf{-2-b=-3}−2−b=−3
So
\mathbf{b=1}b=1 ...9)
10. Now from equation 9), equation 8) and equation 5)
\mathbf{y= ax^{2}+bx+6}y=ax
2
+bx+6
\mathbf{y= -x^{2}+x+6}y=−x
2
+x+6 ...10)
Equation 10) is required quadratic polynomial.
Answer:
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