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Ask QuestionPosted by Prince Raj 5 years, 11 months ago
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Sia ? 5 years, 10 months ago
We have the polynomial f(x) = (3x4 - 15x3 + 13x2 + 25x - 30).
Since √53 and −√53 are the zeros of f(x),
i.e., x=−√53 and x=√53
i.e., x+√53=0andx−√53=0 so it follows that each
one of (x−√53)and(x+√53) is a factor of f(x).so using (a+b)(a−b)=a2−b2, we get
∴(x−√5√3)(x+√5√3)=0⇒(x2−53)=0⇒(3x2−5)3=0 is also a factor of f(x).
Consequently, (3x2 - 5) is a factor of f{x).
On dividing the polynomial by (3x2 - 5), we get
∴ f(x) = 3x4 - 15x3 + 13x2 + 25x - 30
= (3x2 - 5 )(x2 - 5x + 6).
By middle term factorisation. We get,
=(3x2−5)(x2−2x−3x+6)
=([(√3x)2−(√5)2][x(x−2)−3(x−2)]
By using a2−b2=(a+b)(a−b)we get,
=(√3x+√5)(√3x−√5)(x−2)(x−3)
∴ f(x) = 0 , so either factors can equated to zero to get the roots ⇒(√3x+√5)=0 or (√3x−√5)=0
⇒(x - 2) = 0 or (x - 3) = 0
⇒x=−√53 or x=√53 or x = 2 or x = 3
Hence, we get all zeros of f(x) are √53,−√53, 2 and 3.
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Sia ? 5 years, 10 months ago
Since (3y−1),(3y+5) and (5y+1) are in AP, we have
(3y+5)−(3y−1)=(5y+1)−(3y+5)
⇒ 3y+5−3y+1=5y+1−3y−5
⇒ 6=2y−4
⇒ 2y=10
⇒ y=5
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Gaurav Seth 5 years, 11 months ago
A polynomial looks like this:
example of a polynomial this one has 3 terms |
Polynomial comes from <i>poly-</i> (meaning "many") and <i>-nomial</i> (in this case meaning "term") ... so it says "many terms"
A polynomial can have:
constants (like 3, −20, or ½) |
variables (like <i>x</i> and <i>y</i>) |
exponents (like the 2 in y2), but only 0, 1, 2, 3, ... etc are allowed |
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Sia ? 5 years, 10 months ago
Let the speed of train =x km/h and let the speed of bus =y km/h
According to given conditions,
60x+240y=4and100x+200y=4+1060
Let 1x=pand1y=q
Putting this in the above equations, we get
60p +240q =4 ........................... (1)
And100p +200q = 256 ............................. (2)
Multiplying (1) by 5 and (2) by 3, we get
300p +1200q = 20 ......................... (3)
300p +600q = 252......................... (4)
Subtracting (4) from (3), we get
600q =20− 252 =7.5 ⇒ q = 7.5600
Putting value of q in (2), we get
100p +200(7.5600)=256
⇒ 100p + 2.5 = 256
⇒ 100p = 256 - 2.5
⇒ p = 10600
But 1x=pand1y=q
Therefore, x = 60010=60 km/h and y = 6007.5 = 80km/h
Therefore, speed of train =60 km/h
And, speed of bus =80 km/h
0Thank You