If (\alpha,\beta) are roots of (x^2-5x+1=0), what is (\alpha^4+\beta^4)?
Correct answer: A
Here (\alpha+\beta=5) and (\alpha\beta=1). First (\alpha^2+\beta^2=23), then (\alpha^4+\beta^4=23^2-2=527).
Mathematics
Roots of a Quadratic Equation
द्विघात समीकरण के मूल
Roots of a Quadratic Equation, taught in Class 10 Mathematics under the chapter Quadratic Equations, introduces the values of the variable that make a quadratic expression equal to zero. Students learn to identify roots, verify them by substitution, and connect the sum and product of the roots with the coefficients. The topic also helps them form a quadratic equation when its roots are known and use these relationships to solve and check mathematical problems.
TOPIC PRACTICE
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Expert · Level 6
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- (527)
- (529)
- (531)
- (533)
Expert · Level 6
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- (33)
- (45)
- (49)
- (94)
Expert · Level 6
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- (-16)
- (16)
- (-8)
- (8)
Expert · Level 6
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- (4)
- (\frac{7}{2})
- (\frac{9}{2})
- (5)
Expert · Level 6
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- (4)
- (\frac{7}{2})
- (\frac{9}{2})
- (5)
Expert · Level 6
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- \(x^2-2x-1=0\)
- \(x^2-2x+1=0\)
- \(x^2-5x+6=0\)
- \(x^2+2x+5=0\)
Expert · Level 6
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- (0<\lambda<36)
- (\lambda=36)
- (\lambda>36)
- (\lambda<0)
Expert · Level 6
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- (-12)
- (-11)
- (-10)
- (-9)
Expert · Level 6
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- (-13)
- (-11)
- (-9)
- (-7)
Expert · Level 6
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- (100)
- (84)
- (64)
- (42)
Expert · Level 6
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- 31
- 33
- 35
- 41
Expert · Level 6
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- \(h\)
- \(\frac{h}{4}\)
- \(4h\)
- \(h+1\)
Expert · Level 6
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- Not real for every real (a)
- Equal real for every real (a)
- Two real distinct for every real (a)
- One root is always (0)
Expert · Level 6
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- (1) and (5)
- (2) and (4)
- (3) and (3)
- (0) and (6)
Expert · Level 6
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- -211
- -213
- -215
- -217
Expert · Level 6
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- (5)
- (3)
- (1)
- (2a+1)
If both roots of (x^2-14x+m=0) are prime numbers, what is the sum of possible values of (m)?
Correct answer: D
The prime pairs with sum (14) are ((3,11)) and ((7,7)). Thus (m=33) or (m=49), and the sum is (82), so none of the options is correct.
If (x^2+px+64=0) has equal and positive roots, what is the value of (p)?
Correct answer: A
For equal roots, (p^2-256=0), so (p=\pm16). The equal root (-\frac{p}{2}) must be positive, hence (p=-16).
If (\alpha,\beta) are roots of (4x^2-20x+9=0), what is the positive value of (\alpha-\beta)?
Correct answer: B
Here (\alpha+\beta=5) and (\alpha\beta=\frac{9}{4}). Thus ((\alpha-\beta)^2=25-9=16), so the positive difference is (4); option (A) should be correct.
If (\alpha,\beta) are roots of (4x^2-20x+9=0), what is the correct positive value of (\alpha-\beta)?
Correct answer: A
Here (\alpha+\beta=5) and (\alpha\beta=\frac{9}{4}). Since ((\alpha-\beta)^2=25-9=16), the positive difference is (4).
Which of the following quadratic equations has real, distinct, and irrational roots?
Correct answer: A
For \(x^2-2x-1=0\), the discriminant is \(D=b^2-4ac=4+4=8\). Since \(D>0\), the roots are real and distinct; since 8 is not a perfect square, they are irrational. Exam tip: check both the sign and square nature of \(D\).
For (x^2+12x+\lambda=0) to have real distinct roots and both negative roots, what is the correct condition on (\lambda)?
Correct answer: A
For both roots to be negative, the sum (-12) and product (\lambda>0) are needed. For real distinct roots, (144-4\lambda>0), so (0<\lambda<36).
If (\alpha,\beta) are roots of (x^2-7x+10=0), what is (\alpha^2-6\alpha+\beta^2-6\beta)?
Correct answer: B
Here (\alpha+\beta=7) and (\alpha\beta=10). Since (\alpha^2+\beta^2=49-20=29), the value is (29-6(7)=-13), so none of the options is correct.
If (\alpha,\beta) are roots of (x^2-7x+10=0), what is the correct value of (\alpha^2-6\alpha+\beta^2-6\beta)?
Correct answer: A
(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=49-20=29). Therefore the value is (29-6(\alpha+\beta)=29-42=-13).
If (\alpha,\beta) are roots of (x^2+10x+21=0), what is ((\alpha-\beta)^2+4\alpha\beta)?
Correct answer: A
We know ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). Hence ((\alpha-\beta)^2+4\alpha\beta=(\alpha+\beta)^2=100).
If the roots of \\(x^2+px+q=0\\) are \\(-4\\) and \\(9\\), what is the value of \\(p-q\\)?
Correct answer: A
By Vieta’s formulas, the sum of the roots equals \\(-p\\), while their product equals \\(q\\). Thus, \\((-4)+9=5=-p\\), so \\(p=-5\\). Also, \\((-4)(9)=-36\\), giving \\(q=-36\\). Therefore, \\(p-q=-5-(-36)=31\\). Exam tip: For \\(x^2+px+q=0\\), the sum of the roots is \\(-p\\) and the product is \\(q\\).
If one root of the equation \(4x^2-(4h+1)x+h=0\) is \(\frac{1}{4}\), what is the other root?
Correct answer: A
For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\frac{c}{a}\). Here, \(a=4\) and \(c=h\), so the product of the roots is \(\frac{h}{4}\). If one root is \(\frac{1}{4}\), the other root is \(\frac{h/4}{1/4}=h\). Option B is only the product of the roots, not the other root. Exam tip: Divide the product of the roots by the given root to find the second root.
What is the nature of the roots of (x^2-2x+(a^2+3)=0)?
Correct answer: A
The discriminant is (D=4-4(a^2+3)=-4a^2-8). It is negative for every real (a), so the roots are not real.
If (\alpha,\beta) are roots of (x^2-6x+c=0) and (\alpha^2+\beta^2=26), what are the roots?
Correct answer: A
Here (\alpha+\beta=6) and (\alpha^2+\beta^2=26). From (36-2\alpha\beta=26), (\alpha\beta=5), so the roots are (1) and (5).
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2+x-6=0\), what is the value of \(\alpha^5+\beta^5\)?
Correct answer: A
The equation \(x^2+x-6=0\) factors as \((x-2)(x+3)=0\), so its roots are \(2\) and \(-3\). Hence, \(\alpha^5+\beta^5=2^5+(-3)^5=32-243=-211\). Therefore, option A is correct. Exam tip: For a power sum such as this, factor the quadratic first when the roots are integral, and remember that an odd power preserves the negative sign of a negative root.
If (\alpha,\beta) are the roots of (x^2-(2a+1)x+a^2+a-6=0), what is the positive value of (\alpha-\beta)?
Correct answer: A
Here (\alpha+\beta=2a+1) and (\alpha\beta=a^2+a-6). Since ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=25), the positive difference is (5).
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