If one root of (x^2-5x+q=0) is (2), what will be the other root?
Correct answer: A
The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
View question detailsMathematics
Methods of Solving Quadratic Equations
द्विघात समीकरणों को हल करने की विधियाँ
In this Class 10 Mathematics topic from the chapter Quadratic Equations, students learn how to find the values of an unknown variable that satisfy a quadratic equation. They practise solving equations by factorisation, completing the square, and using the quadratic formula, while learning when each method is useful. The topic also develops skills in identifying coefficients, calculating the discriminant, checking solutions, and interpreting whether an equation has two, one, or no real roots.
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Hard · Level 34 · quadratic,roots,sum-of-roots
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- (3)
- (2)
- (5)
- (-3)
Hard · Level 34 · quadratic equations,methods of solving,roots and coefficients,product of roots
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- 6
- 10
- 3
- 5
Hard · Level 34 · quadratic,error-checking,factorisation
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- Correct
- Incorrect because the second root is (2)
- Incorrect because the first root is (-\frac{1}{2})
- Incorrect because both roots are positive
Hard · Level 34 · quadratic,perfect-square,irrational
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- ((x+\sqrt{3})^2=0)
- ((x-\sqrt{3})^2=0)
- ((x+3)^2=0)
- ((x+2\sqrt{3})^2=0)
Hard · Level 34 · quadratic,repeated-root,irrational
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- (x=-\sqrt{3})
- (x=\sqrt{3})
- (x=-3)
- (x=3)
Hard · Level 34 · quadratic,symbolic,difference-of-squares
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- (x=a+b,a-b)
- (x=-a+b,-a-b)
- (x=b+a,b-a)
- (x=a^2,b^2)
Hard · Level 34 · quadratic,simplification,symbolic
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- (x=a,b)
- (x=-a,-b)
- (x=4a,4b)
- (x=a+b,ab)
Hard · Level 34 · quadratic,substitution,biquadratic
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- (y^2-5y+4=0)
- (y^2+5y+4=0)
- (y^2-5x+4=0)
- (y-5y+4=0)
Hard · Level 34 · quadratic,substitution,real-solutions
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- (x=\pm1,\pm2)
- (x=1,2)
- (x=-1,-2)
- (x=\pm4,\pm5)
Hard · Level 34 · quadratic,reciprocal-equation,standard-form
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- (2x^2-5x+2=0)
- (2x^2+5x+2=0)
- (x^2-5x+2=0)
- (2x^2-5=0)
Hard · Level 34 · quadratic,reciprocal-equation,solutions
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- (x=2,\frac{1}{2})
- (x=-2,-\frac{1}{2})
- (x=5,2)
- (x=\frac{5}{2},1)
Hard · Level 34 · quadratic,rational-equation,standard-form
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- (x^2+2x-5=0)
- (x^2-2x-5=0)
- (x^2+2x+5=0)
- (x^2+x-6=0)
Hard · Level 34 · quadratic,rational-equation,correction
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- (x^2-4x+1=0)
- (x^2+2x-5=0)
- (x^2-6x+1=0)
- (x^2+4x+1=0)
Hard · Level 34 · quadratic,quadratic-formula,irrational-roots
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- (x=2\pm\sqrt{3})
- (x=-2\pm\sqrt{3})
- (x=4\pm\sqrt{3})
- (x=2\pm3)
Hard · Level 34 · quadratic,sum-of-roots,parameter
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- (15)
- (-15)
- (5)
- (36)
Hard · Level 34 · quadratic,product-of-roots,parameter
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- (6)
- (3)
- ( \frac{3}{8}) / (\frac{3}{8})
- ( \frac{7}{2}) / (\frac{7}{2})
Hard · Level 34 · quadratic,roots-expression,hard
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- (61)
- (121)
- (30)
- (91)
Hard · Level 34 · quadratic,roots-expression,reciprocal
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- ( \frac{7}{10}) / (\frac{7}{10})
- ( \frac{10}{7}) / (\frac{10}{7})
- (7)
- (10)
Hard · Level 34 · quadratic,sum-of-roots,formula
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- ( \frac{9}{2}) / (\frac{9}{2})
- ( \frac{4}{2}) / (\frac{4}{2})
- ( -\frac{9}{2}) / (-\frac{9}{2})
- (2)
Hard · Level 34 · quadratic,product-of-roots,formula
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- (2)
- ( \frac{9}{2}) / (\frac{9}{2})
- (4)
- ( -2) / (-2)
If one root of the quadratic equation \(x^2-5x+q=0\) is 2, what is the value of \(q\)?
Correct answer: A
Since 2 is a root, substitute \(x=2\) in the equation: \(2^2-5(2)+q=0\), giving \(4-10+q=0\) and hence \(q=6\). Alternatively, the sum of the roots is 5, so the other root is 3 and their product is \(q=2\times3=6\). Exam tip: for \(x^2+bx+c=0\), the product of the roots is \(c\); 10 is related to neither the required constant term nor the correct substitution result.
View question detailsA student wrote (x=\frac{1}{2},-2) for (2x^2+3x-2=0). How is this answer?
Correct answer: A
(2x^2+3x-2=(2x-1)(x+2)), so (x=\frac{1}{2},-2) is correct. In exams, change signs carefully from factors.
View question detailsIn which form can (x^2+2\sqrt{3}x+3=0) be written?
Correct answer: A
Since (3=(\sqrt{3})^2) and the middle term is (2\sqrt{3}x), it is ((x+\sqrt{3})^2). In exams, identify perfect squares even with irrational coefficients.
View question detailsWhat is the root of (x^2+2\sqrt{3}x+3=0)?
Correct answer: A
((x+\sqrt{3})^2=0), so the repeated root is (-\sqrt{3}). In exams, ((x+a)^2=0) gives (x=-a).
View question detailsIf (x^2-2ax+a^2-b^2=0), what are its roots?
Correct answer: A
It is ((x-a)^2-b^2=0), so (x-a=\pm b) and (x=a\pm b). In exams, use the difference of squares.
View question detailsAfter simplifying (4x^2-4(a+b)x+4ab=0), what will be the roots?
Correct answer: A
Dividing the whole equation by (4) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first is easier.
View question detailsIf (y=x^2) is put in (x^4-5x^2+4=0), which equation is obtained?
Correct answer: A
Since (x^4=(x^2)^2=y^2), the new equation is (y^2-5y+4=0). In exams, substitution can simplify difficult forms.
View question detailsWhat are the real solutions of (x^4-5x^2+4=0)?
Correct answer: A
From (y^2-5y+4=0), (y=1,4), so (x^2=1,4) and (x=\pm1,\pm2). In exams, do not forget to return to (x).
View question detailsFor (\frac{1}{x}+x=\frac{5}{2}), (x\neq0), what quadratic form is obtained?
Correct answer: A
Multiplying both sides by (2x) gives (2+2x^2=5x), that is (2x^2-5x+2=0). In exams, remember the condition (x\neq0).
View question detailsWhat are the solutions of (\frac{1}{x}+x=\frac{5}{2}), (x\neq0)?
Correct answer: A
(2x^2-5x+2=(2x-1)(x-2)), so (x=\frac{1}{2}) and (2). In exams, check whether obtained roots are valid in the original equation.
View question detailsFor (\frac{x+1}{x}=\frac{6}{x+1}), (x\neq0,-1), what quadratic form is obtained?
Correct answer: A
Cross multiplication gives ((x+1)^2=6x), so (x^2+2x+1=6x), and the correct form is (x^2-4x+1=0). In exams, cross multiply very carefully.
View question detailsWhat is the correct quadratic form of (\frac{x+1}{x}=\frac{6}{x+1}), (x\neq0,-1)?
Correct answer: A
From ((x+1)^2=6x), we get (x^2+2x+1-6x=0), that is (x^2-4x+1=0). In exams, avoid a wrong middle term.
View question detailsWhat are the roots of (x^2-4x+1=0) by quadratic formula?
Correct answer: A
(D=(-4)^2-4(1)(1)=12), so (x=\frac{4\pm2\sqrt{3}}{2}=2\pm\sqrt{3}). In exams, simplify the square root.
View question detailsIf the sum of roots of (3x^2+px+12=0) is (-5), what is (p)?
Correct answer: A
The sum of roots is (-\frac{p}{3}), so (-\frac{p}{3}=-5) gives (p=15). In exams, remember the sum formula (-\frac{b}{a}).
View question detailsIf the product of roots of (4x^2-7x+p=0) is (\frac{3}{2}), what is (p)?
Correct answer: A
The product of roots is (\frac{p}{4}), so (\frac{p}{4}=\frac{3}{2}) gives (p=6). In exams, use the product formula (\frac{c}{a}).
View question detailsIf the roots of (x^2-11x+30=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?
Correct answer: A
(\alpha+\beta=11) and (\alpha\beta=30), so (\alpha^2+\beta^2=(11)^2-2(30)=61). In exams, remember the identity ((\alpha+\beta)^2-2\alpha\beta).
View question detailsIf the roots of (x^2-7x+10=0) are (\alpha) and (\beta), what is (\frac{1}{\alpha}+\frac{1}{\beta})?
Correct answer: A
(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}). In exams, first write sum and product for reciprocal questions.
View question detailsIf (\alpha,\beta) are roots of (2x^2-9x+4=0), what is (\alpha+\beta)?
Correct answer: A
The sum of roots is (-\frac{b}{a}=-\frac{-9}{2}=\frac{9}{2}). In exams, keep the sign of (b) carefully.
View question detailsIf (\alpha,\beta) are roots of (2x^2-9x+4=0), what is (\alpha\beta)?
Correct answer: A
The product of roots is (\frac{c}{a}=\frac{4}{2}=2). In exams, use (\frac{c}{a}) for the product.
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