Roots of a Quadratic Equation

Correct answer: A. (527)

Explanation: Here (\alpha+\beta=5) and (\alpha\beta=1). First (\alpha^2+\beta^2=23), then (\alpha^4+\beta^4=23^2-2=527).

Correct answer: D. (94)

Explanation: 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.

Correct answer: A. (-16)

Explanation: For equal roots, (p^2-256=0), so (p=\pm16). The equal root (-\frac{p}{2}) must be positive, hence (p=-16).

Correct answer: B. (\frac{7}{2})

Explanation: 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.

Correct answer: A. (4)

Explanation: Here (\alpha+\beta=5) and (\alpha\beta=\frac{9}{4}). Since ((\alpha-\beta)^2=25-9=16), the positive difference is (4).

Correct answer: A. \(x^2-2x-1=0\)

Explanation: 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 \(D\).

Correct answer: A. (0<\lambda<36)

Explanation: 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).

Correct answer: B. (-11)

Explanation: 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.

Correct answer: A. (-13)

Explanation: (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=49-20=29). Therefore the value is (29-6(\alpha+\beta)=29-42=-13).

Correct answer: A. (100)

Explanation: We know ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). Hence ((\alpha-\beta)^2+4\alpha\beta=(\alpha+\beta)^2=100).

Correct answer: A. 31

Explanation: 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\\).

Correct answer: A. \(h\)

Explanation: 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.

Correct answer: A. Not real for every real (a)

Explanation: 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.

Correct answer: A. (1) and (5)

Explanation: 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).

Correct answer: A. -211

Explanation: 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.

Correct answer: A. (5)

Explanation: 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).