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Introduction to Quadratic Equations, part of the Class 10 Mathematics chapter Quadratic Equations, helps students recognise equations of degree two and write them in the standard form ax² + bx + c = 0, where a ≠ 0. Students learn the meaning of coefficients, variables, and constants, identify quadratic equations from examples, and understand how their roots or solutions relate to the equation. The topic builds a foundation for solving quadratic equations by methods such as factorisation and applying the quadratic formula.
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Hard · Level 30 · quadratic-equations,roots-of-quadratic, vieta-formulas,expression-valueView options
0
10
4
28
Hard · Level 30 · quadratic-equations,standard-form,algebraic-expansion,polynomial-simplificationView options
2x^2+10x-54=0)
2x^2+10x+6=0)
x^2+5x-54=0)
2x^2+5x-60=0)
Hard · Level 30 · quadratic-equations,nature-of-roots,discriminant,real-rootsView options
It has no real roots
It has two equal real roots
It has two distinct real roots
It has one positive and one negative real root
Hard · Level 30 · quadratic-equations,roots-and-coefficients,product-of-roots,vieta-formulas,one-rootView options
m
4m
m+4
m-4
Hard · Level 30 · quadratic-equations,roots,discriminant,root-differenceView options
\(\frac{1}{2}\)
\(\frac{3}{2}\)
\(2\)
\(\frac{5}{2}\)
Hard · Level 30 · quadratic-equations,perfect-square,discriminant,parameterView options
49
14
28
196
Hard · Level 30 · quadratic-equations,equal-roots,roots-and-coefficients,vieta-formulasView options
14
-14
7
-7
Hard · Level 30 · quadratic-equations,sum-product,signs,hardView options
(x^2+7x+12=0)
(x^2-7x+12=0)
(x^2+7x-12=0)
(x^2-7x-12=0)
Hard · Level 30 · quadratic equations, degree of polynomial, algebra, class 10 mathematics, equation classificationView options
\(x^2-5x+6=0\)
\(x^3-5x+6=0\)
\(\frac{1}{x}+5=0\)
\(2x-7=0\)
Hard · Level 30 · quadratic-equations,forming-equation,sum-product,hardView options
(x^2-13x+40=0)
(x^2+13x+40=0)
(x^2-40x+13=0)
(x^2+40x-13=0)
Hard · Level 30 · quadratic-equations,reciprocal-roots,hardView options
( \frac{7}{12} )
( \frac{12}{7} )
(7)
(12)
Hard · Level 30 · quadratic-equations,equal-roots,negative-roots,discriminant,perfect-squareView options
x² + 16x + 64 = 0
x² − 16x + 64 = 0
x² − 64 = 0
x² + 64 = 0
Hard · Level 30 · quadratic-equations,algebraic-identities,coefficient-comparison,polynomial-expansionView options
-6
6
-12
36
Hard · Level 30 · quadratic-equations,discriminant,perfect-square,nature-of-rootsView options
Two equal real roots
Two distinct real roots
No real roots
The nature of the roots cannot be determined
Hard · Level 30 · quadratic-equations,reciprocal-roots,conceptual,hardView options
It is not possible
(m=25)
(m=-25)
(m=0)
Hard · Level 30 · quadratic-equations,equal-positive-roots,parameter,hardView options
(-8)
(8)
(4)
(-4)
Hard · Level 30 · quadratic-equations,not-root,parameter,hardView options
(k\neq1)
(k=1)
(k\neq3)
(k=3)
Hard · Level 30 · quadratic-equations,sum-product,forming-equation,hardView options
(x^2-81=0)
(x^2+81=0)
(x^2+9x-81=0)
(x^2-9x-81=0)
Hard · Level 30 · quadratic-equations,roots,parameter,hardView options
(6)
(-6)
(12)
(-12)
Hard · Level 30 · quadratic-equations,roots,cubes-sum,hardView options
(117)
(27)
(90)
(-117)
Question 1HardLevel 30
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2+7x+10=0\), what is the value of \((\alpha+2)(\beta+2)\)?
Correct answer: A
By Vieta’s formulas, \(\alpha+\beta=-\frac{7}{1}=-7\) and \(\alpha\beta=\frac{10}{1}=10\). Therefore, \((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=10+2(-7)+4=0\). Option B is only the value of \(\alpha\beta\), not of the complete expression. Exam tip: For expressions involving roots, first find the sum and product of the roots using Vieta’s formulas.
What is the standard form ax^2+bx+c=0) of the equation (x+2)(x+5)+(x-1)(x+4)=609?
Correct answer: A
Expand the two products: (x+2)(x+5)=x^2+7x+10) and (x-1)(x+4)=x^2+3x-4). Their sum is 2x^2+10x+6). Bringing 60 to the left gives 2x^2+10x+6-60=0), which simplifies to 2x^2+10x-54=0). Option B stops before subtracting 60, while option C does not represent the correctly simplified equation. Exam tip: In standard form, collect all terms on one side and write zero on the other side.
Which of the following statements is correct for the equation \\(x^2-6x+11=0\\)?
Correct answer: A
For this quadratic equation, \(a=1\), \(b=-6\), and \(c=11\). Its discriminant is \(D=b^2-4ac=(-6)^2-4(1)(11)=36-44=-8\). Since \(D<0\), the equation has no real roots. Exam tip: \(D=0\) gives equal real roots, while \(D>0\) gives two distinct real roots.
If one root of the equation \(x^2-(m+4)x+4m=0\) is \(4\), what is the other root?
Correct answer: A
For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, the product of the roots is \(4m\). Since one root is \(4\), the other root is \(\frac{4m}{4}=m\). Remember that \(m+4\) is the sum of the roots, not the other root.
What is the difference between the roots of the equation \(2x^2-9x+10=0\), taking the larger root minus the smaller root?
Correct answer: A
Here, the discriminant is \(D=b^2-4ac=(-9)^2-4(2)(10)=81-80=1\). The difference between the roots is \(\frac{\sqrt{D}}{|a|}\), so it equals \(\frac{\sqrt{1}}{2}=\frac{1}{2}\). Alternatively, \(2x^2-9x+10=(2x-5)(x-2)\), giving roots \(\frac{5}{2}\) and \(2\), whose difference is \(\frac{1}{2}\). Exam tip: use \(\frac{\sqrt{D}}{|a|}\) when only the difference between roots is required.
If \(x^2-14x+c=0\) is a perfect-square quadratic equation, what is the value of \(c\)?
Correct answer: A
For the quadratic to be a perfect square, its left-hand polynomial must be expressible as \(x^2-14x+c=(x-7)^2\). Since \((x-7)^2=x^2-14x+49\), we get \(c=49\). Exam tip: for \(x^2+bx+c\) to be a perfect square, \(c=\left(\frac{b}{2}\right)^2\); here, \(b=-14\).
If the two roots of the quadratic equation \(x^2+bx+49=0\) are equal and each root is \(-7\), what is the value of \(b\)?
Correct answer: A
The sum of the roots is \((-7)+(-7)=-14\). For a quadratic equation \(x^2+bx+c=0\), the sum of the roots is \(-b\). Thus, \(-b=-14\), giving \(b=14\). Option B is the sum of the roots, not the value of the coefficient \(b\). Exam tip: Use the relation \(\text{sum of roots}=-\frac{\text{coefficient of }x}{\text{coefficient of }x^2}\).
Which of the following is a quadratic equation in one variable?
Correct answer: A
A quadratic equation has the form \(ax^2+bx+c=0\), where \(a\ne0\). In option A, the highest power of x is 2. Option B is cubic, while D is linear. Exam tip: check the degree after simplifying fractions.
Which of the following quadratic equations will have two equal and negative roots?
Correct answer: A
For option A, \(x^2+16x+64=(x+8)^2\). Thus, \((x+8)^2=0\) gives both roots as \(x=-8\), so they are equal and negative. In option B, \((x-8)^2=0\), giving equal but positive roots. Exam tip: For equal roots, first check that the discriminant is zero, then determine the sign of the root.
If \\(x+a\\)^2 = x^2 - 12x + 36 holds for all real x, what is the value of a?
Correct answer: A
Expanding gives \\(x+a\\)^2 = x^2 + 2ax + a^2\\). Comparing the coefficients of x, \\(2a=-12\\), so \\(a=-6\\). The constant term gives \\(a^2=36\\), which alone allows both 6 and -6; the coefficient of x identifies -6 uniquely. Exam tip: For an identity, compare coefficients of like powers of the variable.
If \(k\) is a real constant, what is the nature of the roots of the equation \(9x^2+12kx+4k^2=0\)?
Correct answer: A
The equation can be rewritten as \((3x+2k)^2=0\). Thus \(3x+2k=0\), giving the repeated root \(x=-\frac{2k}{3}\). Equivalently, the discriminant is \(D=(12k)^2-4(9)(4k^2)=0\), so the roots are equal and real. Exam tip: for a quadratic equation, \(D=0\) indicates two equal real roots.
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