Question

Maths students please answer all questions correctly. Correct answer I will choose for brainliest answer. You can answer for graphs by capturing photo of book after drawing graph in it.

1. Find the zeroes of polynomial p(x) = 3x2+15x.

2.Find roots of -21x2 +16x +5.

3. Find roots of p(x) = x3 – 4x. Also represent it’s graph.

4. Divide 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3.

5. One zero of the polynomial p(x) = x3 + 6x2 + 11x + 6 is (-2), find other zeroes of the polynomial.

6. A person buys five tickets of ahmedabad to delhi and ten tickets of ahmedabad to mumbai in Rs 1050.If total fare of one ticket each is Rs 130, find individual ticket price.

7. In ΔABC m∠ A = x, m∠ B = 3x and m∠ C =y. If 3y-5x =30. Show that ABC is a right angle triangle.Also find values of all angles.

8. A linear equation is given as 4x + 7y = 28.
Write another linear equation in two variables such that the geometrical representation of the pair so formed is
(i) Intersecting lines.
(ii) Parallel lines.
(iii)Coinciding lines.
Also draw graph for all lines you find.

Answer 1

here is your answer u r looking for..

see the attachment

Answer 2

HELLO !!!

GOOD MORNING,

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●●(1) ●●
$3 {x}^{2} + 15x = 0 \\ \\ = > 3x(x + 5) = 0 \\ \\ = > 3x = 0 \: \: \: or \: \: \: \: x + 5 = 0 \\ \\ = > x = 0 \: \: \: \: or \: \: \: \: x = - 5$

●So , zeroes of the polynomial p (x) are 0 and -5.


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●●(2) ●●
$- 21 {x}^{2} + 16x + 5 = 0 \\ \\ = > - 21 {x}^{2} + 21x - 5x + 5 = 0 \\ \\ = > 21x( - x + 1) + 5( - x + 1) = 0 \\ \\ = > (21x + 5) \: \: (1 - x) = 0 \\ \\ = > 21x + 5 = 0 \: \: \: \: or \: \: \: \: 1 - x = 0 \\ \\ = > x = \frac{ - 5}{21} \: \: \: \: or \: \: \: \: x = 1$

● So, the zeroes are -5/12 and 1 .



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●●(3) ●●

We have,

p(x) = x^3 − 4x

Now zeroes of p(x) are given by solving the equation


${x}^{3} - 4x = 0 \\ \\ = > x( {x}^{2} - 4) = 0 \\ \\ = > x(x + 2)(x -2 ) = 0\\ \\ = > x = 0 \: \: \: or \: \: \: x = - 2 \: \: \: or \: \: \: x = 2$


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●●(4) ●●

Let's divide 2t^4 + 3t^3 – 2t^2 – 9t – 12 by t^2 – 3.


( Refer attachment for answer )


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●● (5) ●●

Given:  p(x)= x^3−6x^2+11x−6 and its factor, x-2

Let us divide p(x) by (x+2).


( Refer attachment for division )


${x}^{3} + 6 {x}^{2} + 11x + 6 = (x + 2)( {x}^{2} + 4x + 3) \\ \\ \: \: = (x + 2) \: \: ( {x}^{2} + 3x + x + 3) \\ \\ \: \: = (x + 2) \: \: (x(x + 3) + 1(x + 3)) \\ \\ \: \: = (x + 2) \: \: ((x + 3)(x + 1)) \\ \\ \: \: = x = - 2 \: \: or \: \: \: x = - 3 \: \: or \: \: x = - 1$

Hence , the other two zeroes are -3 and -1 .



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●● (7) ●●

We have a triangle ABC , As :

( see in attachment )


And given
3 y - 5x  = 30°               -------------- ( 1 )

And  we know sum of all angles in the triangle = 180°
So,
x + 3x  + y = 180°
4x + y = 180°              ---------------- ( 2 )

Multiply by 3 in equation 2 , and get
12x + 3y = 540°          ---------------- ( 3 )

Now,

equation 3  -  equation 1 , we get

=> 17x = 510°

x  = 30°  Substitute this value in equation 2 , we get

y = 60°

So,

∠ a  = x  = 30°
∠ b = 3x  = 90°
∠ c  = x   = 60°

we can see that ∠ b = 90°  , so this triangle is a right angle triangle               ( Ans )




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Hope it helps u !!!!!


# Nikky