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The Zeroes of the quadratic Polynomial x2
55x + 125 are
Answers
Answer:
just apply the quadratic formula
\begin{gathered}d = {b}^{2} - 4ac \\ and \: then \\ roots = \frac{ - b + - \sqrt{d} }{2a}\end{gathered}
d=b
2
−4ac
andthen
roots=
2a
−b+−
d
OR
The roots are x=-44+\sqrt{1811},-44-\sqrt{1811}x=−44+
1811
,−44−
1811
Step-by-step explanation:
Given : Polynomial - x^2+88x+125x
2
+88x+125
To find : The zero of the quadratic polynomial
Solution :
Equation x^2+88x+125=0x
2
+88x+125=0
Solving by discriminant method
General form - ax^2+bx+c=0ax
2
+bx+c=0
D=b^2-4acD=b
2
−4ac
Solution is x=\frac{-b\pm\sqrt{D}}{2a}x=
2a
−b±
D
Comparing with our equation,
where a=1 , b=88, c=125
D=b^2-4acD=b
2
−4ac
D=(88)^2-4(1)(125)D=(88)
2
−4(1)(125)
D=7744-500D=7744−500
D=7244D=7244
Solution is x=\frac{-b\pm\sqrt{D}}{2a}x=
2a
−b±
D
x=\frac{-(88)\pm\sqrt{7244}}{2(1)}x=
2(1)
−(88)±
7244
x=\frac{-88\pm2\sqrt{1811}}{2}x=
2
−88±2
1811
x=-44\pm\sqrt{1811}x=−44±
1811
SIMILAR QUESTION BY UNDERSTANDING THIS FORMULA TRY TO SOLVE THAT QUESTIONThe Zeroes of the quadratic Polynomial x2
55x + 125 are