if the roots of the quadratic equation hx² + 21x + 10 = 0 where h≠0 are in the ratio 2:5 , find the possible value of h ?
answer it please
Answers
Answer:
Solution:
Roots are in ratio = 2 : 5
Let the roots be 2x and 5x
Then,
\begin{gathered}sum\ of\ roots = 2x + 5x = 7x \\\\product\ of\ roots = 2x \times 5x = 10x^2\end{gathered}
sum of roots=2x+5x=7x
product of roots=2x×5x=10x
2
We know,
\begin{gathered}ax^2+bx+c = 0\\\\sum\ of\ zeros = \frac{-b}{a} \\\\product\ of\ zeros = \frac{c}{a}\end{gathered}
ax
2
+bx+c=0
sum of zeros=
a
−b
product of zeros=
a
c
Therefore,
From given,
hx^2 + 21x + 10 = 0hx
2
+21x+10=0
\begin{gathered}sum\ of\ zeros = \frac{-21}{h} \\\\product\ of\ zeros = \frac{10}{h}\end{gathered}
sum of zeros=
h
−21
product of zeros=
h
10
Already we found,
\begin{gathered}sum\ of\ roots = 7x \\\\product\ of\ roots = 10x^2\end{gathered}
sum of roots=7x
product of roots=10x
2
Thus,
\begin{gathered}7x = \frac{-21}{h} \\\\ x = \frac{-3}{h} \\\\\end{gathered}
7x=
h
−21
x=
h
−3
Compare product of zeros,
\begin{gathered}10x^2 = \frac{10}{h}\\\\Substitute\ x = \frac{-3}{h}\\\\10 \times (\frac{-3}{h})^2 = \frac{10}{h} \\\\\frac{9}{h^2} = \frac{1}{h}\\\\h = 9\end{gathered}
10x
2
=
h
10
Substitute x=
h
−3
10×(
h
−3
)
2
=
h
10
h
2
9
=
h
1
h=9
Then h is 9