If (\alpha,\beta) are roots of (4x^2-13x+3=0), what is (\alpha+\beta)?
Correct answer: A
The sum of roots is (-\frac{b}{a}=-\frac{-13}{4}=\frac{13}{4}). In exams, keep the sign of (b) carefully.
Mathematics
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.
Hard · Level 6 · 19 questions
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Hard · Level 6
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- ( \frac{13}{4}) / (\frac{13}{4})
- ( \frac{3}{4}) / (\frac{3}{4})
- ( -\frac{13}{4}) / (-\frac{13}{4})
- (4)
Hard · Level 6
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- ( \frac{3}{4}) / (\frac{3}{4})
- ( \frac{13}{4}) / (\frac{13}{4})
- (3)
- (-\frac{3}{4})
Hard · Level 6
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- \( \frac{121}{16}\) / \(\frac{121}{16}\)
- \( \frac{169}{16}\) / \(\frac{169}{16}\)
- (3)
- \( \frac{73}{16}\) / \(\frac{73}{16}\)
Hard · Level 6
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- (n>25)
- (n<25)
- (n=25)
- (n\le25)
Hard · Level 6
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- (n<25)
- (n>25)
- (n=25)
- (n\ge25)
Hard · Level 6
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- \((5x+3)(x-2)=0\)
- \((5x-3)(x+2)=0\)
- \((x+3)(5x-2)=0\)
- \((5x+2)(x-3)=0\)
Hard · Level 6
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- \(x=2,\,-\frac{3}{5}\)
- \(x=-2,\,\frac{3}{5}\)
- \(x=1,\,-6\)
- \(x=\frac{7}{5},\,-1\)
Hard · Level 6
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- \(x=-4\pm\sqrt{11}\)
- \(x=4\pm\sqrt{11}\)
- \(x=-8\pm\sqrt{11}\)
- \(x=-4\pm11\)
Hard · Level 6
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- (-96)
- (96)
- (-16)
- (6)
Hard · Level 6
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- ( \frac{15\sqrt{2}}{2}) / (\frac{15\sqrt{2}}{2})
- (-\frac{15\sqrt{2}}{2})
- (10\sqrt{2})
- (15)
Hard · Level 6
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- (\sqrt{37})
- (37)
- (7)
- (3)
Hard · Level 6
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- There are no real roots
- There are two equal real roots
- There are two distinct real roots
- One root is (0)
Hard · Level 6
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- (4(x-1)^2+5=0)
- (4(x+1)^2+5=0)
- (4(x-1)^2-5=0)
- (4(x-2)^2+9=0)
Hard · Level 6
View options
- (x^2-16x+55=0)
- (x^2-10x+16=0)
- (x^2-4x+7=0)
- (x^2+16x+55=0)
Hard · Level 6
View options
- ( \frac{106}{45}) / (\frac{106}{45})
- ( \frac{196}{45}) / (\frac{196}{45})
- ( \frac{14}{45}) / (\frac{14}{45})
- ( \frac{45}{106}) / (\frac{45}{106})
Hard · Level 6
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- 1
- 2
- 7
- 49
Hard · Level 6
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- (k=-2)
- (k=2)
- (k=-4)
- (k=4)
Hard · Level 6
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- (x^2-13x+22=0)
- (x^2-13x+50=0)
- (x^2+13x+22=0)
- (x^2-5x+36=0)
Hard · Level 6
View options
- (x=\frac{13\pm9}{2})
- (x=\frac{-13\pm9}{2})
- (x=13\pm9)
- (x=\frac{13\pm\sqrt{22}}{2})
If (\alpha,\beta) are roots of (4x^2-13x+3=0), what is (\alpha\beta)?
Correct answer: A
The product of roots is (\frac{c}{a}=\frac{3}{4}). In exams, use (\frac{c}{a}) for the product.
If \(\alpha,\beta\) are roots of \(4x^2-13x+3=0\), what is \((\alpha-\beta)^2\)?
Correct answer: A
\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{13}{4}\right)^2-3=\frac{121}{16}\). In exams, use this identity for the square of difference.
If (x^2-10x+n=0) has no real roots, which condition on (n) is correct?
Correct answer: A
For no real roots, (D<0), so (100-4n<0) and (n>25). In exams, connect (D<0) with no real roots.
If (x^2-10x+n=0) has two distinct real roots, which condition on (n) is correct?
Correct answer: A
For two distinct real roots, (D>0), so (100-4n>0) and (n<25). In exams, connect (D>0) with distinct real roots.
Which of the following is the correct factorised form for solving the equation \(5x^2-7x-6=0\)?
Correct answer: A
Expanding option A gives \((5x+3)(x-2)=5x^2-10x+3x-6=5x^2-7x-6\), so it is the correct factorised form. Option B produces a middle coefficient of \(+7\), while options C and D produce \(+13\) and \(-13\), respectively. In an exam, always expand the factors and compare the result with the original quadratic equation.
What are the roots of the quadratic equation \(5x^2-7x-6=0\)?
Correct answer: A
The quadratic factors as \(5x^2-7x-6=(5x+3)(x-2)\). Thus, \((5x+3)(x-2)=0\) gives \(x=-\frac{3}{5}\) or \(x=2\), so option A is correct. In option B, the signs of both roots are incorrect. Exam tip: when setting each factor equal to zero, remember to change the sign while solving for the root.
What are the roots of the equation \(x^2+8x+5=0\)?
Correct answer: A
Completing the square in \(x^2+8x+5=0\) gives \(x^2+8x+16=11\), or \((x+4)^2=11\). Hence, \(x+4=\pm\sqrt{11}\), so the roots are \(x=-4\pm\sqrt{11}\). Option B has the wrong sign before 4, while option D incorrectly omits the square-root sign. Exam tip: the \(\pm\) symbol represents both roots, so write both values when required.
If the roots of (x^2-6x-16=0) are (\alpha,\beta), what is (\alpha^2\beta+\alpha\beta^2)?
Correct answer: A
(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=6) and (\alpha\beta=-16), so the value is (-96). In exams, factor the expression first.
If one root of (x^2+px+25=0) is double the other and both are negative, what is (p)?
Correct answer: A
Let the roots be (-r) and (-2r), then (2r^2=25) and (p=3r=\frac{15\sqrt{2}}{2}). In exams, do not forget to rationalize the denominator.
What is the difference between the roots of (x^2-7x+3=0)?
Correct answer: A
Here (D=(-7)^2-4(1)(3)=37), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{37}). In exams, the difference of roots can be found directly from (D).
What is the correct statement about real roots of (4x^2-8x+9=0)?
Correct answer: A
Here (D=(-8)^2-4(4)(9)=-80<0), so there are no real roots. In exams, (D<0) means no real roots.
Which completed square form is correct for (4x^2-8x+9=0)?
Correct answer: A
(4x^2-8x+9=4(x-1)^2+5), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
If roots of (x^2-10x+16=0) are (\alpha,\beta), which equation has roots (\alpha+3,\beta+3)?
Correct answer: A
The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.
If the roots of (x^2-14x+45=0) are (\alpha,\beta), what is (\frac{\alpha}{\beta}+\frac{\beta}{\alpha})?
Correct answer: A
(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}), where (\alpha+\beta=14) and (\alpha\beta=45), so the value is (\frac{196-90}{45}=\frac{106}{45}). In exams, convert expressions into sum and product.
If \(kx^2-14x+49=0\) has equal roots, what is the value of \(k\)?
Correct answer: A
For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-14\), and \(c=49\), so \(D=(-14)^2-4(k)(49)=196-196k\). Thus, \(196-196k=0\), giving \(k=1\). Exam tip: identify the coefficient of \(x^2\) as \(a=k\); the equation is quadratic only when \(k\neq 0\).
What is (k) for equal roots of (x^2-2(k+4)x+k^2=0)?
Correct answer: A
(D=4(k+4)^2-4k^2=0) gives ((k+4)^2=k^2), so (8k+16=0) and (k=-2). In exams, expand squares carefully.
If ((x-4)(x-9)=14), what is the standard quadratic equation?
Correct answer: A
((x-4)(x-9)=x^2-13x+36), so (x^2-13x+36=14) gives (x^2-13x+22=0). In exams, bring all terms to one side after expansion.
What are the roots of (x^2-13x+22=0) by quadratic formula?
Correct answer: A
Here (D=(-13)^2-4(1)(22)=81), so (x=\frac{13\pm9}{2}). In exams, if (D) is a perfect square, the answer simplifies quickly.
QUIZ COMPLETE