Roots of a Quadratic Equation

Let the two equal roots be r and r. Since the equation is monic, the product of the roots is the constant term, so r^2=25. The condition that both roots are negative selects r=-5 rather than r=5. Their sum is therefore r+r=-10. For x^2+px+25=0, the sum of the roots equals -p by Vieta’s relation. Hence -p=-10, which gives p=10. Equivalently, the equation becomes (x+5)^2=x^2+10x+25, confirming the value. Option B would produce roots 5 and 5, which are positive, while 5 and -5 do not form equal roots and do not have product 25 as a repeated pair.

Putting (x=2) gives (4+3a=0), so (a=-\frac{4}{3}). The product is (a), so the other root is (-\frac{2}{3}).

For the quadratic equation x² - 7x + q = 0, let the roots be α and β. By the relationships between roots and coefficients, α + β = 7 and αβ = q. The difference is given as 1, so take α - β = 1; if the order is reversed, the absolute difference is still 1. Solving α + β = 7 and α - β = 1 gives 2α = 8, hence α = 4, and β = 3. Therefore q = αβ = 4 × 3 = 12. Equivalently, the discriminant is (α - β)² = 1, so 49 - 4q = 1, which also gives q = 12. The other options do not satisfy both the sum and the stated root difference.

Here the sum of roots is (S) and product is (P). Therefore ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=S^2-4P).

The given equation factors as \(x^2-6x-16=(x-8)(x+2)\), so its roots are 8 and −2. Adding 1 to each root gives 9 and −1. Their sum is 8 and their product is −9; therefore, the required monic equation is \(x^2-8x-9=0\). Option B retains the old roots, while option C uses an incorrect sum of the new roots. Exam tip: for roots \(\alpha\) and \(\beta\), form the monic equation as \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

The roots of the equation are 8 and -2, since \(x^2-6x-16=(x-8)(x+2)\). After subtracting 2 from each root, the new roots are 6 and -4, whose product is \(6\times(-4)=-24\). Remember that the transformation must be applied to both roots; changing only one root gives an incorrect result.

For the original equation, Vieta’s relations give α+β=4 and αβ=3. If the new roots are 3α and 3β, their sum is 3α+3β=3(α+β)=12, while their product is (3α)(3β)=9αβ=27. A monic quadratic with roots r and s is x^2-(r+s)x+rs=0. Substituting the new sum and product gives x^2-12x+27=0, so option A is correct. Option B keeps the old sum, option C uses the old product multiplied by only 3 instead of 9, and option D has the wrong sign for the x-term. The result can also be checked from the original roots 1 and 3, which become 3 and 9.

By Vieta’s formula, the sum of the roots of 8x^2-6x+5=09 is 8\alpha+\beta9 = -\frac{-6}{1}=6. Hence, 8\alpha+39+8\beta+39 = \alpha+\beta+6 = 6+6=12. The distractor 9 results from adding 3 only once. Exam tip: when the same number is added to both roots, their sum increases by twice that number.

The quadratic equation (x^2-12x+36=0) factors as ((x-6)^2=0). Hence both roots are 6, so (\alpha=\beta=6) and (\alpha-\beta=6-6=0). Exam tip: when the discriminant (D=b^2-4ac) is zero, the two roots are equal and their difference is zero.

The quadratic x^2+4x-21=0 factors as (x+7)(x-3)=0, because -7+3=-4 and (-7)(3)=-21. Thus its roots are 3 and -7. The absolute difference does not depend on which root is called α or β: |α-β|=|3-(-7)|=|10|=10. The same result follows from the discriminant. For a monic quadratic, the squared difference of its roots is Δ=b^2-4ac; here Δ=4^2-4(1)(-21)=16+84=100, so |α-β|=√100=10. Therefore option A is correct. The values 4, 7, and 21 are coefficients or related numbers, but none represents the separation between the two roots.

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta). From (52=36-2\alpha\beta), we get (\alpha\beta=-8).

Let the roots be \(t\) and \(2t\). Their sum is \(9\), so \(t+2t=9\), giving \(t=3\). Hence, the roots are \(3\) and \(6\). The product of the roots equals \(c\), so \(c=3\times6=18\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\).

Since both roots are negative and their ratio is \(1:2\), let them be \(-t\) and \(-2t\). Their product is \((-t)(-2t)=2t^2=18\), giving \(t=3\). Thus, the roots are \(-3\) and \(-6\), whose sum is \(-9\). By Vieta’s relation, the sum of the roots is \(-p\), so \(-p=-9\) and \(p=9\). Exam tip: for \(x^2+px+c=0\), the sum of the roots is \(-p\).

If the roots are \(\alpha\) and \(\beta\), then \(\alpha\beta=\frac{c}{a}\). Reciprocal roots have product \(1\), so \(c=a\). The condition \(b=0\) indicates additive inverse roots instead. Exam tip: use \(c/a\) for the product of roots.

The sum is (-\frac{b}{a}=\frac{3}{4}) and the product is (\frac{k}{4}). From (\frac{k}{4}=\frac{3}{4}), (k=3).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-4\), and \(c=m+4\), so \(D=b^2-4ac=16-4(m+4)=-4m\). Thus, \(-4m<0\), which gives \(m>0\). If \(m=0\), then \(D=0\), giving two equal real roots; therefore, option C is incorrect. Exam tip: For questions about the nature of roots, first determine the sign of the discriminant.

The equation can be rewritten as \((x-(m+2))^2=0\). Hence its repeated root is \(x=m+2\), and the roots are equal for every real value of \(m\). Therefore, option A is correct. Option C gives only the special case \(m=-2\), not the complete set of valid values. Exam tip: in such questions, first look for a perfect square or verify that the discriminant \(D=b^2-4ac\) is zero.

By Vieta’s formulas, \(\alpha+\beta=9\) and \(\alpha\beta=18\). Therefore, \((\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=18-18+4=4\). Hence, option A is correct. Exam tip: 18 is only the product \(\alpha\beta\); the required expression also involves the sum of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, \(a=5\) and \(c=q\), so the product of the roots is \(\frac{q}{5}\). Therefore, \(\frac{q}{5}=3\times(-4)=-12\). The distractor \(12\) results from missing the negative sign. Exam tip: remember that the sum of roots is \(-\frac{b}{a}\), while their product is \(\frac{c}{a}\).

The sum of the roots is \(5+(-2)=3\). Since the sum of roots is \(-p\), we get \(p=-3\). Their product is \(q=5\times(-2)=-10\). Therefore, \(p-q=-3-(-10)=7\). Exam tip: For \(x^2+px+q=0\), the sum of roots is \(-p\) and their product is \(q\).