If one root of (x^2-23x+q=0) is (9), what will be the other root?
Correct answer: A
The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum 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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Expert · Level 35 · quadratic,roots,sum-of-roots
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- (14)
- (9)
- (23)
- (-14)
Expert · Level 35 · quadratic,parameter,product-of-roots
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- (126)
- (207)
- (14)
- (23)
Expert · Level 35 · quadratic,error-checking,factorisation
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- Correct
- Incorrect because the second root is (\frac{3}{5})
- Incorrect because the first root is (-\frac{1}{2})
- Incorrect because both roots are positive
Expert · Level 35 · quadratic,perfect-square,irrational
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- ((x+\sqrt{13})^2=0)
- ((x-\sqrt{13})^2=0)
- ((x+13)^2=0)
- ((x+2\sqrt{13})^2=0)
Expert · Level 35 · quadratic,repeated-root,irrational
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- (x=-\sqrt{13})
- (x=\sqrt{13})
- (x=-13)
- (x=13)
Expert · Level 35 · quadratic,symbolic,factorisation
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- (x=s,t)
- (x=-s,-t)
- (x=s+t,st)
- (x=s-t,t-s)
Expert · Level 35 · quadratic,symbolic,difference-of-squares
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- (x=r+s,r-s)
- (x=-r+s,-r-s)
- (x=s+r,s-r)
- (x=r^2,s^2)
Expert · Level 35 · quadratic,simplification,symbolic
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- (x=m,n)
- (x=-m,-n)
- (x=36m,36n)
- (x=m+n,mn)
Expert · Level 35 · quadratic,substitution,biquadratic
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- (y^2-20y+64=0)
- (y^2+20y+64=0)
- (y^2-20x+64=0)
- (y-20y+64=0)
Expert · Level 35 · quadratic,substitution,real-solutions
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- (x=\pm2\sqrt{2},\pm2\sqrt{2})
- (x=\pm4,\pm2)
- (x=\pm8,\pm20)
- (x=8,12)
Expert · Level 35 · quadratic,reciprocal-equation,standard-form
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- (6x^2-37x+6=0)
- (6x^2+37x+6=0)
- (x^2-37x+6=0)
- (6x^2-37=0)
Expert · Level 35 · quadratic,reciprocal-equation,solutions
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- (x=6,\frac{1}{6})
- (x=-6,-\frac{1}{6})
- (x=37,6)
- (x=\frac{37}{6},1)
Expert · Level 35 · quadratic,rational-equation,standard-form
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- (x^2-26x+25=0)
- (x^2+10x-11=0)
- (x^2-36x+25=0)
- (x^2+26x+25=0)
Expert · Level 35 · quadratic,rational-equation,solutions
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- (x=1,25)
- (x=-1,-25)
- (x=5,36)
- (x=26,25)
Expert · Level 35 · quadratic,quadratic-formula,perfect-square-discriminant
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- (x=1,11)
- (x=-1,-11)
- (x=6\pm5)
- (x=12\pm\sqrt{11})
Expert · Level 35 · quadratic,sum-of-roots,parameter
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- (72)
- (-72)
- (9)
- (256)
Expert · Level 35 · quadratic,product-of-roots,parameter
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- (3)
- (9)
- ( \frac{1}{27}) / (\frac{1}{27})
- ( \frac{20}{3}) / (\frac{20}{3})
Expert · Level 35 · quadratic,roots-expression,expert
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- (277)
- (529)
- (126)
- (403)
Expert · Level 35 · quadratic,roots-expression,reciprocal
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- ( \frac{20}{91}) / (\frac{20}{91})
- ( \frac{91}{20}) / (\frac{91}{20})
- (20)
- (91)
Expert · Level 35 · quadratic,sum-of-roots,formula
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- ( \frac{25}{7}) / (\frac{25}{7})
- ( \frac{12}{7}) / (\frac{12}{7})
- ( -\frac{25}{7}) / (-\frac{25}{7})
- (7)
If one root of (x^2-23x+q=0) is (9), what is the value of (q)?
Correct answer: A
The other root is (14), so (q=9\times14=126). In exams, when (a=1), the constant term is the product of roots.
View question detailsA student wrote (x=\frac{1}{2},-\frac{3}{5}) for (10x^2+x-3=0). How is this answer?
Correct answer: A
(10x^2+x-3=(5x+3)(2x-1)), so (x=\frac{1}{2},-\frac{3}{5}) is correct. In exams, change signs carefully from factors.
View question detailsIn which form can (x^2+2\sqrt{13}x+13=0) be written?
Correct answer: A
Since (13=(\sqrt{13})^2) and the middle term is (2\sqrt{13}x), it is ((x+\sqrt{13})^2). In exams, identify perfect squares even with irrational coefficients.
View question detailsWhat is the root of (x^2+2\sqrt{13}x+13=0)?
Correct answer: A
((x+\sqrt{13})^2=0), so the repeated root is (-\sqrt{13}). In exams, ((x+a)^2=0) gives (x=-a).
View question detailsWhat are the roots of (x^2-(s+t)x+st=0)?
Correct answer: A
The equation is equivalent to ((x-s)(x-t)=0), so the roots are (s) and (t). In exams, apply zero product rule to symbolic factors too.
View question detailsIf (x^2-2rx+r^2-s^2=0), what are its roots?
Correct answer: A
It is ((x-r)^2-s^2=0), so (x-r=\pm s) and (x=r\pm s). In exams, quickly recognize the difference of squares.
View question detailsAfter simplifying (36x^2-36(m+n)x+36mn=0), what will be the roots?
Correct answer: A
Dividing the whole equation by (36) gives (x^2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.
View question detailsIf (y=x^2) is put in (x^4-20x^2+64=0), which equation is obtained?
Correct answer: A
Since (x^4=(x^2)^2=y^2), the new equation is (y^2-20y+64=0). In exams, use substitution to form a quadratic.
View question detailsWhat are the real solutions of (x^4-20x^2+64=0)?
Correct answer: B
From (y^2-20y+64=0), (y=4,16), so (x^2=4,16) and (x=\pm2,\pm4). In exams, do not forget to return to (x).
View question detailsFor (\frac{1}{x}+x=\frac{37}{6}), (x\neq0), what quadratic form is obtained?
Correct answer: A
Multiplying both sides by (6x) gives (6+6x^2=37x), that is (6x^2-37x+6=0). In exams, remember the condition (x\neq0).
View question detailsWhat are the solutions of (\frac{1}{x}+x=\frac{37}{6}), (x\neq0)?
Correct answer: A
(6x^2-37x+6=(6x-1)(x-6)), so (x=\frac{1}{6}) and (6). In exams, check whether obtained roots are valid in the original equation.
View question detailsWhat is the correct quadratic form of (\frac{x+5}{x}=\frac{36}{x+5}), (x\neq0,-5)?
Correct answer: A
Cross multiplication gives ((x+5)^2=36x), so (x^2+10x+25-36x=0), and (x^2-26x+25=0). In exams, cross multiply carefully.
View question detailsWhat are the solutions of (\frac{x+5}{x}=\frac{36}{x+5}), (x\neq0,-5)?
Correct answer: A
(x^2-26x+25=(x-1)(x-25)), so (x=1) and (x=25). In exams, check solutions against excluded denominator values.
View question detailsWhat are the roots of (x^2-12x+11=0) by quadratic formula?
Correct answer: A
(D=(-12)^2-4(1)(11)=100), so (x=\frac{12\pm10}{2}) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.
View question detailsIf the sum of roots of (8x^2+px+32=0) is (-9), what is (p)?
Correct answer: A
The sum of roots is (-\frac{p}{8}), so (-\frac{p}{8}=-9) gives (p=72). In exams, remember the sum formula (-\frac{b}{a}).
View question detailsIf the product of roots of (9x^2-20x+p=0) is (\frac{1}{3}), what is (p)?
Correct answer: A
The product of roots is (\frac{p}{9}), so (\frac{p}{9}=\frac{1}{3}) gives (p=3). In exams, use the product formula (\frac{c}{a}).
View question detailsIf the roots of (x^2-23x+126=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?
Correct answer: A
(\alpha+\beta=23) and (\alpha\beta=126), so (\alpha^2+\beta^2=23^2-2(126)=277). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).
View question detailsIf the roots of (x^2-20x+91=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{20}{91}). In exams, first write sum and product in reciprocal questions.
View question detailsIf (\alpha,\beta) are roots of (7x^2-25x+12=0), what is (\alpha+\beta)?
Correct answer: A
The sum of roots is (-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}). In exams, keep the sign of (b) carefully.
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