Find its zeroes:
3√2x^2 + 13 x+6√2
Answers
Answer:
Let us first factorize the given equation 3
2
x
2
+13x+6
2
as shown below:
⇒3
2
x
2
+13x+6
2
=3
2
x
2
+9x+4x+6
2
=3x(
2
x+3)+2
2
(
2
x+3)
=(3x+2
2
)(
2
x+3)
Therefore, the zeroes of the given polynomial are:
(3x+2
2
)=0
⇒x=−
3
2
2
=α
(
2
x+3)=0
⇒x=−
2
3
=β
Now, the sum and product of the roots is as follows:
α+β=−
3
2
2
−
2
3
=−(
3
2
2
+
2
3
)=
3
2
−4−9
=−
3
2
13
=−
a
b
αβ=(−
3
2
2
)(−
2
3
)=2=
3
2
6
2
=
a
c
Hence verified.
Step-by-step explanation:
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Step-by-step explanation:
p(x) = 3√2x² + 13x + 6√2
p(x) = 3√2x² +9x + 4x + 6√2
p(x) = 3√2x² + 9x + 2√2*√2x + 6√2
p(x) = 3x (√2x + 3) + 2√2(√2x +3)
p(x) = (√2x + 3) (3x + 2√2)
to find zeros. equate p(x) to zero and find the values of 'x' which are the zeros
therefore (√2x + 3) = 0 => x = -3/√2
3x + 2√2 = 0 => x = -2√2/3