Roots of a Quadratic Equation

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.

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

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 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).

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.

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

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

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\\).

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.

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.

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).

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.

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).

Factor out the common factor x from the equation: x² − 16x = x(x − 16) = 0. The zero-product property states that if a product is zero, at least one factor must be zero. Therefore x = 0 or x − 16 = 0. Solving the second factor gives x = 16, so the two roots are 0 and 16. Hence option A is correct. A common error is to read the coefficient −16 as a root and write −16; however, the factor is x − 16, whose zero is positive 16. Options C and D introduce values not obtained from either factor. Substitution confirms the result: both 0² − 16(0) and 16² − 16(16) equal zero.

The governing principle is that dividing an equation by a variable factor can discard a solution when that factor itself may be zero. The original equation factors as x² − 5x = x(x − 5) = 0. By the zero-product property, either x = 0 or x − 5 = 0, which gives x = 5. If we divide by x immediately, we implicitly assume x is not zero and obtain only x − 5 = 0. Thus the root x = 0 is lost. Option A is correct. The root x = 5 is retained, while −5 and 1 do not satisfy the original equation. The safe method is to factor first and consider every factor separately.

The governing concept is factorisation followed by the zero-product property. We seek two factors of 7x² + 8x + 1. It factors as (7x + 1)(x + 1), because their product is 7x² + 7x + x + 1 = 7x² + 8x + 1. Hence (7x + 1)(x + 1) = 0, so either 7x + 1 = 0 or x + 1 = 0. These give x = −1/7 and x = −1. Therefore option A is correct. Options B and D lose the negative signs, while option C treats the coefficient 7 as though it were a root rather than solving 7x + 1 = 0.