Examine the nature of the roots of the equations: (i) 2x^2 + 2x + 3 = 0 -Maths 9th

Examine the nature of the roots of the equations: (i) 2x^2 + 2x + 3 = 0 -Maths 9th
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A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

Of the following quadratic equations, which is the one whose roots are 2 and – 15 ? -Maths 9th

Description : Of the following quadratic equations, which is the one whose roots are 2 and – 15 ? -Maths 9th

Last Answer : answer:

Explain Nature of Roots of a quadratic equation. -Maths 9th

Description : Explain Nature of Roots of a quadratic equation. -Maths 9th

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Determine the nature of the roots of the following equatins : (a) `x^(3) + 2x +4 = 0` (b) `3x^(2) - 10x + 3 = 0` (c ) `x^(2) - 24x + 144 = 0`

Description : Determine the nature of the roots of the following equatins : (a) `x^(3) + 2x +4 = 0` (b) `3x^(2) - 10x + 3 = 0` (c ) `x^(2) - 24x + 144 = 0`

Last Answer : Determine the nature of the roots of the following equatins : (a) `x^(3) + 2x +4 = 0` (b) `3x^(2) - 10x + 3 = 0` (c ) `x^(2) - 24x + 144 = 0`

Examine , whether the following numbers are rational or irrational : -Maths 9th

Description : Examine , whether the following numbers are rational or irrational : -Maths 9th

Last Answer : ∴ It is an irrational number .

Examine whether the following numbers are rational or irrational: -Maths 9th

Description : Examine whether the following numbers are rational or irrational: -Maths 9th

Last Answer : 1 irrational no. 2 rational no. 3 irrational no.

Examine , whether the following numbers are rational or irrational : -Maths 9th

Description : Examine , whether the following numbers are rational or irrational : -Maths 9th

Last Answer : ∴ It is an irrational number .

Examine whether the following numbers are rational or irrational: -Maths 9th

Description : Examine whether the following numbers are rational or irrational: -Maths 9th

Last Answer : 1 irrational no. 2 rational no. 3 irrational no.

Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Description : Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Last Answer : Let the common root be α ⟹α2−kα−21=0......(1) α2−3kα+35=0........(2) (1)−(2)⟹2kα=56⟹α=k28​Substituting α=k28​ in (1) (k28​)2−28−21=0 ⟹k=±4

What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Description : What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Last Answer : (c) (8, 6)Let AB be the given line 4x + 3y = 25 Let O′(a, b) be the image of O in the given line AB. Let O O′ cut AB in point P. Also OP ⊥ AB and P is the mid-point of OO′. ∴ Co-ordinates of P are \(\bigg( ... 4 imes6}{3}\) = 8∴ The image of the point O(0, 0) in the line 4x + 3y - 25 = 0 is (8, 6).

If the roots of the equations ax2 + 2bx + c = 0 and bx2 – are simultanously real then prove that b2 = 4ac. -Maths 10th

Description : If the roots of the equations ax2 + 2bx + c = 0 and bx2 – are simultanously real then prove that b2 = 4ac. -Maths 10th

Last Answer : Let D1​ and D2​ be the discriminants of the two equations, then D1​≥0 and D2​≥0 ⇒4b2−4ac≥0 and 4ac−4b2≥0 ⇒b2≥ac and ac≥b2 ⇒b2=ac

graph of 2x+9=0 in one variable -Maths 9th

Description : graph of 2x+9=0 in one variable -Maths 9th

Last Answer : 2x+9=0 2x=-9 x=-9/2 x=-4.5

If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form -Maths 9th

Description : Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form -Maths 9th

Last Answer : (a) The given linear equation is 2x + 0y + 9 = 0 ⇒ 2x + 9 = 0 ⇒ 2x = -9 ⇒ x= - 9/2 and y can be any real number. Hence, (-9/2 , m) is the required form of solution of the given linear equation.

graph of 2x+9=0 in one variable -Maths 9th

Description : graph of 2x+9=0 in one variable -Maths 9th

Last Answer : 2x+9=0 2x=-9 x=-9/2 x=-4.5

If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form -Maths 9th

Description : Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form -Maths 9th

Last Answer : (a) The given linear equation is 2x + 0y + 9 = 0 ⇒ 2x + 9 = 0 ⇒ 2x = -9 ⇒ x= - 9/2 and y can be any real number. Hence, (-9/2 , m) is the required form of solution of the given linear equation.

express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Description : express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Last Answer : NEED ANSWER

express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Description : express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Last Answer : Let y=0. So, the equation will be 2x+0y-5=0. Then from the equation, a=2; b=0; c=-5

If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Description : If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

If 2x^2 – 7xy + 3y^2 = 0, then the value of x : y is -Maths 9th

Description : If 2x^2 – 7xy + 3y^2 = 0, then the value of x : y is -Maths 9th

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For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

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If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th

Description : If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th

Last Answer : answer:

Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th

Description : Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th

Last Answer : Let A(1, 2) and B(11, 9) be the given points. Let the points of trisection be P and Q. Then,AP = PQ = QB = k (say)⇒ AQ = AP + PQ = 2k and PB = PQ + QB = 2k ∴ AP : PB = k : 2k = 1 : 2 and AQ ... two points of trisection are \(\big(rac{13}{3},rac{13}{3}\big)\) and \(\big(rac{23}{3},rac{20}{3}\big)\).

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4

Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

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Last Answer : this is the correct answer!

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Description : Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

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Last Answer : Following are the rational numbers which represent irrational numbers .

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Description : How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Last Answer : (c) Let the linear equation be ax + by + c = 0. On putting x = 1 and y = 2, in above equation we get =s a + 2b + c = 0, where a, b and c, are real number Here, different values of a, b and ... a + 2b + c = 0. Hence, infinitely many linear equations in x and yean be satisfied by x = 1 and y = 2.

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Description : Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Description : In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Last Answer : Following are the rational numbers which represent irrational numbers .

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Description : How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Last Answer : (c) Let the linear equation be ax + by + c = 0. On putting x = 1 and y = 2, in above equation we get =s a + 2b + c = 0, where a, b and c, are real number Here, different values of a, b and ... a + 2b + c = 0. Hence, infinitely many linear equations in x and yean be satisfied by x = 1 and y = 2.

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

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What is distance between the graphs of the equations y = -1 and y = 3? -Maths 9th

Description : What is distance between the graphs of the equations y = -1 and y = 3? -Maths 9th

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How many linear equations satisfy x = 2 and y = -3? -Maths 9th

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Write three possible linear equations which can pass through point (3, –2). -Maths 9th

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Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

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Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

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Last Answer : (b) \(\bigg(rac{10}{3},rac{20}{3}\bigg)\). (+ 10, 20) log100 |x+y| = \(rac{1}{2}\) ⇒ |x + y| = 100\(^{rac{1}{2}}\)⇒ |x + y| = 10 as (-10 is inadmissible) ...(i) log10y - log10| x | = log1004⇒ log10 ... x < 0, then x = 10.∴ If x = \(rac{10}{3}\), then y = \(rac{20}{3}\) and if x = 10, y = 20.

If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Description : If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Last Answer : x2+b2=1−2bx ...... (i) x2+b2+2bx=1 (x+b)2=1 x+b= 1 ∴x=1−b,−1−b x2+a2=1−2ax ...... (ii) x2+a2+2ax=1 (x+a)2=1 ∴x=1−a,−1−a It is given that the equations have only one root in ... or −1−a=−b−1 we get a=b but then both roots will be common, which is not possible. Hence, options A,B and C are correct.

Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Description : Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Last Answer : answer: