find the roots of the following question
Answers
Step-by-step explanation:
{5x}^{2} - 29x + 205x
2
−29x+20
Factorise The Term By Middle Term Splitting.
In Middle Term Splitting Divide Middle Term Such that On Multiplying Both It should be Product of Coffecient x² and constant Term
\begin{gathered} {5x}^{2} - 29x + 20 = 0 \\ \\ {5x}^{2} - 25x - 4x + 20 = 0 \\ \\ 5x(x - 5) - 4(x - 5) = 0 \\ \\ (5x - 4)(x - 5) = 0\end{gathered}
5x
2
−29x+20=0
5x
2
−25x−4x+20=0
5x(x−5)−4(x−5)=0
(5x−4)(x−5)=0
Here One Zero of Polynomial
\begin{gathered}5x - 4 = 0 \\ \\ 5x = 4 \\ \\ x = \frac{4}{5} \end{gathered}
5x−4=0
5x=4
x=
5
4
Other Zero of Polynomial
\begin{gathered}x - 5 = 0 \\ \\ x = 5\end{gathered}
x−5=0
x=5
We got Zeroes
\begin{gathered} \alpha = \frac{4}{5} \\ \\ \beta = 5\end{gathered}
α=
5
4
β=5
Polynomial
ax²+bx+c
{5x}^{2} - 29x + 205x
2
−29x+20
Coffecient of x²=5
Coffecient of x¹=-29
Coffecient of x^0=20
Let a in ax²+bx+c=5
Let B in ax²+bx+c=-29
Let c in ax²+bx+c=20
Sum Of Zeroes
\begin{gathered} \alpha + \beta = - \frac{coffecient \: of \: x}{coffecient \: of \: {x}^{2} } \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \frac{4}{5} + 5 = - \frac{-29}{5} \\ \\ \frac{29}{4} = + \frac{29}{4} < /p > < p > \end{gathered}
α+β=−
coffecientofx
2
coffecientofx
α+β=−
a
b
5
4
+5=−
5
−29
4
29
=+
4
29
</p><p>
lhs = rhslhs=rhs
Product of Zeroes
\begin{gathered} \alpha \times \beta = \frac{constant \: term}{cofficient \: of \: {x}^{2} } \\ \\ \frac{4}{5} \times 5 = \frac{20}{5} \\ \\ \frac{20}{5} = \frac{20}{5} \\ \\ lhs = rhs\end{gathered}
α×β=
cofficientofx
2
constantterm
5
4
×5=
5
20
5
20
=
5
20
lhs=rhs
\boxed{\mathbf{\huge{LHS=RHS}}}
LHS=RHS
\underline{\mathbf{\huge{Verified✓}}}
Verified✓
Answer:
5 and 4/5
Step-by-step explanation:
5x² - 29x + 20
=》 5x² -25x -4x + 20
=》 5x(x - 5) -4(x - 5)
=》 (5x - 4) (x - 5)
=》 x=4/5 and x = 5