What roots are obtained for (2x^2+8x+1=0) by completing the square method?
Correct answer: A
Since ((x+2)^2=\frac{7}{2}), (x=-2\pm\sqrt{\frac{7}{2}}=-2\pm\frac{\sqrt{14}}{2}). In exams, write the square root in simplified form.
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,completing-square,irrational-roots
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- (x=-2\pm\frac{\sqrt{14}}{2})
- (x=2\pm\frac{\sqrt{14}}{2})
- (x=-4\pm\sqrt{14})
- (x=-2\pm\sqrt{7})
Hard · Level 35 · quadratic,hard,factorisation
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- (x=\frac{1}{4},\frac{3}{2})
- (x=-\frac{1}{4},-\frac{3}{2})
- (x=\frac{4}{1},\frac{2}{3})
- (x=3,\frac{1}{8})
Hard · Level 35 · quadratic,middle-term-splitting,ac-method
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- (15x^2+10x+6x+4=0)
- (15x^2+12x+4x+4=0)
- (15x^2+15x+x+4=0)
- (15x^2+20x-4x+4=0)
Hard · Level 35 · quadratic,roots,factorisation
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- (x=-\frac{2}{3},-\frac{2}{5})
- (x=\frac{2}{3},\frac{2}{5})
- (x=-\frac{3}{2},-\frac{5}{2})
- (x=-4,-\frac{1}{15})
Hard · Level 35 · quadratic,discriminant,parameter
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- (k=\frac{5}{2})
- (k=-\frac{5}{2})
- (k=2)
- (k=1)
Hard · Level 35 · quadratic,roots-parameter,sum-of-roots
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- (p=10)
- (p=6)
- (p=12)
- (p=4)
Hard · Level 35 · quadratic,completing-square,hard
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- \(\left(x-\frac{9}{5}\right)^2=\frac{36}{25}\)
- \(\left(x+\frac{9}{5}\right)^2=\frac{36}{25}\)
- \(\left(x-\frac{18}{5}\right)^2=\frac{9}{5}\)
- \(\left(x-\frac{9}{5}\right)^2=\frac{81}{25}\)
Hard · Level 35 · quadratic,roots,verification
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- (x=3,\frac{3}{5})
- (x=-3,-\frac{3}{5})
- (x=\frac{5}{3},3)
- (x=1,\frac{9}{5})
Hard · Level 35 · quadratic,discriminant,equal-roots
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- (m=49)
- (m=14)
- (m=28)
- (m=196)
Hard · Level 35 · quadratic,common-root,hard
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- (x=2)
- (x=\frac{4}{3})
- (x=1)
- (x=\frac{1}{2})
Hard · Level 35 · quadratic,roots,sum-of-roots
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- (5)
- (4)
- (9)
- (-5)
Hard · Level 35 · quadratic,parameter,product-of-roots
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- (20)
- (36)
- (5)
- (9)
Hard · Level 35 · quadratic,error-checking,factorisation
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- Correct
- Incorrect because the second root is (1)
- Incorrect because the first root is (-\frac{2}{3})
- Incorrect because both roots are positive
Hard · Level 35 · quadratic,perfect-square,irrational
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- ((x-\sqrt{5})^2=0)
- ((x+\sqrt{5})^2=0)
- ((x-5)^2=0)
- ((x-2\sqrt{5})^2=0)
Medium · Level 35 · quadratic equations,perfect square,repeated root,Methods of Solving Quadratic Equations,Mathematics,Class 10 MCQ
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- x = √5
- x = −√5
- x = 5
- x = −5
Hard · Level 35 · quadratic,symbolic,factorisation
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- (x=m,n)
- (x=-m,-n)
- (x=m+n,mn)
- (x=m-n,n-m)
Hard · Level 35 · quadratic,symbolic,difference-of-squares
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- (x=p+q,p-q)
- (x=-p+q,-p-q)
- (x=q+p,q-p)
- (x=p^2,q^2)
Hard · Level 35 · quadratic,simplification,symbolic
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- (x=r,s)
- (x=-r,-s)
- (x=9r,9s)
- (x=r+s,rs)
Hard · Level 35 · quadratic,substitution,biquadratic
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- (y^2-10y+9=0)
- (y^2+10y+9=0)
- (y^2-10x+9=0)
- (y-10y+9=0)
Hard · Level 35 · quadratic,substitution,real-solutions
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- (x=\pm1,\pm3)
- (x=1,3)
- (x=-1,-3)
- (x=\pm9,\pm10)
What will be the roots of (8x^2-14x+3=0) by factorisation method?
Correct answer: A
(8x^2-14x+3=(4x-1)(2x-3)), so the roots are (\frac{1}{4}) and (\frac{3}{2}). In exams, set each linear factor equal to zero.
View question detailsWhat is the correct splitting of the middle term in (15x^2+16x+4=0)?
Correct answer: A
Here (ac=60) and (10+6=16), so (16x) is split as (10x+6x). In exams, check both sum (b) and product (ac).
View question detailsWhat are the roots of (15x^2+16x+4=0)?
Correct answer: A
(15x^2+16x+4=(3x+2)(5x+2)), so the roots are (-\frac{2}{3}) and (-\frac{2}{5}). In exams, write fractional roots in simplest form.
View question detailsIf (x^2-2(k-1)x+k^2-4=0) has equal roots, what is the value of (k)?
Correct answer: A
For equal roots, (D=0), so (4(k-1)^2-4(k^2-4)=0) gives (-2k+5=0). In exams, expand (D) carefully in parameter questions.
View question detailsIf the roots of (3x^2-(p+2)x+12=0) are (2) and (2), what is (p)?
Correct answer: A
The sum of roots is (4), and (\frac{p+2}{3}=4), so (p=10). In exams, find the sum of roots using (-\frac{b}{a}).
View question detailsWhich middle step is correct while solving \(5x^2-18x+9=0\) by completing the square?
Correct answer: A
First \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) is obtained, then \(\left(x-\frac{9}{5}\right)^2=\frac{36}{25}\). In exams, divide by (a) first when \(a\neq1\).
View question detailsWhat will be the roots of (5x^2-18x+9=0)?
Correct answer: A
(5x^2-18x+9=(5x-3)(x-3)), so the roots are (\frac{3}{5}) and (3). In exams, verify the answer quickly by factorisation.
View question detailsIf (x^2-14x+m=0) has real and equal roots, what is the value of (m)?
Correct answer: A
For real and equal roots, (D=0), so (196-4m=0) gives (m=49). In exams, equal roots mean (D=0).
View question detailsWhich root is common to (3x^2-10x+8=0) and (4x^2-12x+8=0)?
Correct answer: A
The roots of the first equation are (2,\frac{4}{3}), and the roots of the second are (2,1). In exams, solve both equations separately for the common root.
View question detailsIf one root of (x^2-9x+q=0) is (4), what will be the other root?
Correct answer: A
The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
View question detailsIf one root of (x^2-9x+q=0) is (4), what is the value of (q)?
Correct answer: A
The other root is (5), so (q=4\times5=20). In exams, (c) equals the product of roots when (a=1).
View question detailsA student wrote (x=\frac{2}{3},-1) for (3x^2+x-2=0). How is this answer?
Correct answer: A
(3x^2+x-2=(3x-2)(x+1)), so (x=\frac{2}{3},-1) is correct. In exams, change signs carefully from factors.
View question detailsIn which form can (x^2-2\sqrt{5}x+5=0) be written?
Correct answer: A
Since (5=(\sqrt{5})^2) and the middle term is (-2\sqrt{5}x), it is ((x-\sqrt{5})^2). In exams, identify perfect squares even with irrational coefficients.
View question detailsWhat is the root of x² − 2√5x + 5 = 0?
Correct answer: A
The governing concept is solving a quadratic equation by recognizing a perfect square. Compare x² − 2√5x + 5 with the identity (x − c)² = x² − 2cx + c². Taking c = √5 gives c² = 5, so the equation becomes x² − 2√5x + 5 = (x − √5)² = 0. A square can equal zero only when its base is zero; hence x − √5 = 0 and x = √5. The root is repeated because both quadratic roots coincide. Therefore option A is correct. Option B has the wrong sign, while C and D confuse the constant 5 with the root and do not satisfy the equation.
View question detailsWhat are the roots of (x^2-(m+n)x+mn=0)?
Correct answer: A
The equation is equivalent to ((x-m)(x-n)=0), so the roots are (m) and (n). In exams, the same rule applies to symbolic factors.
View question detailsIf (x^2-2px+p^2-q^2=0), what are its roots?
Correct answer: A
It is ((x-p)^2-q^2=0), so (x-p=\pm q) and (x=p\pm q). In exams, recognize the difference of squares.
View question detailsAfter simplifying (9x^2-9(r+s)x+9rs=0), what will be the roots?
Correct answer: A
Dividing the whole equation by (9) gives (x^2-(r+s)x+rs=0). In exams, removing the common factor first makes solving easier.
View question detailsIf (y=x^2) is put in (x^4-10x^2+9=0), which equation is obtained?
Correct answer: A
Since (x^4=(x^2)^2=y^2), the new equation is (y^2-10y+9=0). In exams, substitution can simplify a difficult form.
View question detailsWhat are the real solutions of (x^4-10x^2+9=0)?
Correct answer: A
From (y^2-10y+9=0), (y=1,9), so (x^2=1,9) and (x=\pm1,\pm3). In exams, do not forget to return to (x).
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