What is the nature of roots of (x^2+2\sqrt{3}x+3=0)?
Correct answer: A
Here (D=(2\sqrt{3})^2-4\cdot1\cdot3=0), so the roots are equal and real. In exams, be careful while squaring a surd coefficient.
Mathematics
Nature of Roots
मूलों की प्रकृति
In Class 10 Mathematics, Nature of Roots explains how to determine the type and number of solutions of a quadratic equation. Students use the discriminant, b² − 4ac, to identify whether an equation has two distinct real roots, two equal real roots, or no real roots. The topic connects algebraic calculations with the graph of a quadratic function and helps learners interpret equations, compare cases, and solve related problems from the chapter Quadratic Equations.
TOPIC PRACTICE
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Hard · Level 6
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- Two equal real roots
- Two distinct real roots
- No real roots
- Two rational roots
Hard · Level 6
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- Two distinct real roots
- Two equal real roots
- No real roots
- Two zero roots
Hard · Level 6
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- \(m=3\) या \(m=-5\)
- \(m=4\) या \(m=-4\)
- \(m=5\) या \(m=-3\)
- \(m=2\) या \(m=-6\)
Hard · Level 6
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- (r^2<\frac{25}{4}) and (r\neq0)
- (r^2=\frac{25}{4})
- (r^2>\frac{25}{4})
- (r=0)
Hard · Level 6
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- (\lambda>1)
- (\lambda=1)
- (\lambda<1)
- (\lambda\le1)
Hard · Level 6
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- Two equal real roots
- Two distinct real roots
- No real roots
- Two irrational roots
Hard · Level 6
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- It has two distinct real roots
- It has two equal real roots
- It has no real roots
- It has two rational roots
Hard · Level 6
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- (n<2)
- (n=2)
- (n>2)
- (n\le2)
Hard · Level 6
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- (s=0)
- (s=1)
- (s=2)
- (s=4)
Hard · Level 6
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- Two distinct real roots
- Two equal real roots
- No real roots
- Two rational roots
Hard · Level 6
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- (0) times
- (1) time
- (2) times
- (4) times
Hard · Level 6
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- (0) times
- (1) time
- (2) times
- (3) times
Hard · Level 6
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- (0) times
- (1) time
- (2) times
- (5) times
Hard · Level 6
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- True
- False
- True only when (a=0)
- True only when (c=0)
Hard · Level 6
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- True
- False
- True only when (D=1)
- True only when (a=1)
Hard · Level 6
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- Two distinct real roots
- Two equal real roots
- No real roots
- Two equal irrational roots
Hard · Level 6
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- (h=\frac{3}{4})
- (h=\frac{9}{4})
- (h=3)
- (h=9)
Hard · Level 6
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- Two distinct real and irrational roots
- Two equal real roots
- No real roots
- Two rational roots
Hard · Level 6
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- \(k=\pm9\)
- \(k=\pm18\)
- \(k=\pm3\)
- \(k=\pm27\)
Hard · Level 6
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- Two equal real roots
- Two distinct real roots
- No real roots
- Two irrational roots
Hard · Level 6
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- Two equal real roots
- Two distinct real roots
- No real roots
- Only one root is real
Hard · Level 6
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- (p^2=4q)
- (p^2>4q)
- (p^2<4q)
- (p=4q)
Hard · Level 6
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- Two equal real roots
- Two distinct real roots
- No real roots
- Two unequal imaginary roots
Hard · Level 6
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- Two distinct real roots for all real (k)
- Two equal real roots for all real (k)
- No real roots for any real (k)
- Equal roots only when (k=0)
Hard · Level 6
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- (k\le1) and (k\neq0)
- (k<1) only
- (k>1)
- (k=0)
What will be the nature of roots of (x^2+2\sqrt{2}x+1=0)?
Correct answer: A
Here (D=(2\sqrt{2})^2-4\cdot1\cdot1=4), so (D>0). In exams, (D=4) gives two distinct real roots.
If the two roots of the equation \(x^2-2(m+1)x+16=0\) are equal, what are the possible values of \(m\)?
Correct answer: A
For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=-2(m+1)\), and \(c=16\). Thus, \(D=b^2-4ac=[-2(m+1)]^2-64=0\), giving \((m+1)^2=16\). Hence, \(m+1=\pm4\), so \(m=3\) or \(m=-5\). Exam tip: In questions involving equal roots of a quadratic equation, begin by using \(D=0\).
For which (r) will (rx^2-5x+r=0) have two distinct real roots?
Correct answer: A
For two distinct real roots, (D>0), so (25-4r^2>0). Also (r\neq0) is needed because the equation must remain quadratic.
If (x^2+2x+\lambda=0) has no real roots, what is the correct condition for (\lambda)?
Correct answer: A
For no real roots, (D<0), so (4-4\lambda<0) gives (\lambda>1). In exams, keep strict inequality separate from equality.
State the nature of roots of (5x^2+20x+20=0).
Correct answer: A
Here (D=20^2-4\cdot5\cdot20=0), so the roots are equal and real. In exams, removing a common factor does not change the nature.
Which conclusion is correct for (10x^2-3x+2=0)?
Correct answer: C
Here (D=(-3)^2-4\cdot10\cdot2=-71), so it has no real roots. In exams, small (b^2) and large (4ac) can make (D<0).
If (2x^2-4x+n=0) has real and distinct roots, which condition is correct for (n)?
Correct answer: A
For real and distinct roots, (D>0), so (16-8n>0) gives (n<2). In exams, do not forget coefficient (a) in (4ac).
For which (s) will (s x^2+2sx+1=0) have equal roots?
Correct answer: B
For equal roots, (D=0), so ((2s)^2-4s=0) gives (4s(s-1)=0). (s=0) does not make a quadratic, so (s=1).
If (D=-5), which statement about the roots of a quadratic equation is correct?
Correct answer: C
(D=-5) is negative, so there will be no real roots. In exams, (D<0) may also be linked with complex roots.
How many times will the graph of (x^2+4x+8=0) cut the (x)-axis?
Correct answer: A
Here (D=4^2-4\cdot1\cdot8=-16), so there are no real roots. In exams, (D<0) means no cut with the (x)-axis.
How many times will the graph of (x^2-6x+9=0) touch the (x)-axis?
Correct answer: B
Here (D=(-6)^2-4\cdot1\cdot9=0), so the graph touches the (x)-axis once. In exams, (D=0) indicates tangency.
How many times will the graph of (x^2-5x+4=0) cut the (x)-axis?
Correct answer: C
Here (D=(-5)^2-4\cdot1\cdot4=9), so there are two distinct real roots. In exams, (D>0) means two cuts with the (x)-axis.
Statement: If (b^2-4ac<0), then (ax^2+bx+c=0) has no real roots. How is this statement?
Correct answer: A
When (b^2-4ac<0), the discriminant is negative, so real roots do not exist. In exams, this is the standard rule for nature of roots.
Statement: When (D>0), the roots of a quadratic equation are always equal. How is this statement?
Correct answer: B
When (D>0), roots are not equal; they are distinct real roots. In exams, equal roots occur only when (D=0).
What will be the nature of roots of (2x^2+5x-12=0)?
Correct answer: A
Here (D=5^2-4\cdot2\cdot(-12)=121), so there are two distinct real roots. In exams, (D=121) is also a positive perfect square.
If (x^2-3x+h=0) has equal roots, what is the value of (h)?
Correct answer: B
For equal roots, (D=0), so (9-4h=0) and (h=\frac{9}{4}). In exams, the (D=0) method remains valid even with fractional answers.
What is the nature of roots of (8x^2-8x+3=0)?
Correct answer: C
Here (D=(-8)^2-4\cdot8\cdot3=-32), so there are no real roots. In exams, the sign of (D) gives the final decision.
If the roots of the quadratic equation \(3x^2-2kx+27=0\) are equal, what are the possible values of \(k\)?
Correct answer: A
For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=3\), \(b=-2k\), and \(c=27\), so \((-2k)^2-4(3)(27)=0\), giving \(4k^2=324\). Hence \(k^2=81\), and therefore \(k=\pm9\). Thus, option A is correct. Exam tip: For equal roots of a quadratic equation, set the discriminant equal to zero.
What is the nature of roots of \(x^2+\frac{1}{2}x+\frac{1}{16}=0\)?
Correct answer: A
Here \(D=\left(\frac{1}{2}\right)^2-4\cdot1\cdot\frac{1}{16}=0\), so the roots are equal and real. In exams, square the denominator too while squaring a fraction.
How are the roots of \(4x^2+\frac{4}{3}x+\frac{1}{9}=0\)?
Correct answer: A
Here \(D=\left(\frac{4}{3}\right)^2-4\cdot4\cdot\frac{1}{9}=0\), so the roots are equal and real. In exams, calculate (4ac) carefully with fractions.
If (x^2+px+q=0) has two equal real roots, which relation is correct?
Correct answer: A
For equal roots, (D=0), so (p^2-4q=0) gives (p^2=4q). In exams, the formula becomes simpler when (a=1).
If (b^2=4ac) in a quadratic equation, what will be the nature of its roots?
Correct answer: A
When (b^2=4ac), (D=b^2-4ac=0), so the roots are equal and real. In exams, treat this as the discriminant zero condition.
If (x^2-2(k+3)x+(k^2+6k+5)=0), what will be the nature of its roots?
Correct answer: A
Here (D=[-2(k+3)]^2-4(k^2+6k+5)=16), so (D>0). In exams, the sign of (D) after simplification gives the final answer.
If (kx^2-6x+9=0) has real roots, what is the correct condition for (k)?
Correct answer: A
For real roots, (D\ge0), so (36-36k\ge0) gives (k\le1). Also (k\neq0) is required because the equation must remain quadratic.
QUIZ COMPLETE