Introduction to Quadratic Equations

The product is (11) and the sum is \(-\frac{k}{6}\). From \(11=-2\left(-\frac{k}{6}\right)\), we get \(k=33\).

The sum of roots is (7) and the product is (-18). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=343+378=721).

For the monic quadratic \(x^2+bx+c=0\), Vieta's formulas give sum of roots \(\alpha+\beta=-b\) and product \(\alpha\beta=c\). Here \(\alpha+\beta=-6+13=7\), so \(-b=7\) and hence \(b=-7\). The product is \(-6\times13=-78\), so \(c=-78\). Therefore \(b+c=-7+(-78)=-85\). A common error (leading to option C = 71) is to take the product as +78 instead of -78. Exam tip: apply Vieta’s relations directly and watch sign conventions for sum and product of roots in monic quadratics.

Substituting \(a=-4\) gives \(a+4=0\) and \(a^2-16=0\), so the equation reduces to \(0\cdot x^2+0\cdot x+13=0\), i.e. \(13=0\). This is impossible, so the statement is contradictory. Option B is wrong because the quadratic term vanishes, C is wrong because the linear term also vanishes, and D is wrong because the equality is not true for any x. Exam tip: when a parameter is given, first check whether coefficients of highest-degree terms become zero — that indicates a degenerate (constant) equation leading to either contradiction or identity.

Expanding gives left side (2x^2-2x+85) and right side (x^2+6x+9). Subtracting gives (x^2-8x+76=0).

Substitute \(x=-1\): \(1+(6k-3)(-1)+5k=0\) which simplifies to \(1-6k+3+5k=0\), i.e. \(4-k=0\). Therefore \(k=4\). Option B (\(-4\)) often results from a sign error during substitution; the correct simplification yields a positive 4. Exam tip: Always substitute the root directly and simplify terms carefully, watching for sign changes when \(x\) is negative.

For \(x^2-9x+4=0\) we have \(a=1,\;b=-9,\;c=4\). By Vieta, \(\alpha+\beta=-b/a=9\) and \(\alpha\beta=c/a=4\). Therefore \(\alpha+\beta+5\alpha\beta=9+5\times4=29\). Option B (13) is wrong because it uses \(\alpha+\beta+\alpha\beta=9+4\) instead of adding \(5\alpha\beta\); option C (20) equals only \(5\alpha\beta\); option D (36) equals \((\alpha+\beta)(\alpha\beta)=9\times4\). Exam tip: apply Vieta’s formulas quickly to get sum and product of roots for any quadratic \(ax^2+bx+c=0\).

The direct answer is option A: p = -3. In a quadratic x² + px + q = 0, the sum of the roots is -p because the coefficient of x is p. The stated roots are 6 and p, so their sum is 6 + p. Equating these gives 6 + p = -p. Add p to both sides: 6 + 2p = 0. Therefore 2p = -6 and p = -3. Hence option A is correct. Option B, p = 3, would make the root sum 9, while the coefficient rule gives -3, so it is inconsistent. Option C, p = 6, would make the two roots both 6 and their sum 12, but the rule gives -6. Option D, p = -6, would make the sum 0, while the rule gives 6. The value of q is not needed because the question asks only for p and the sum-of-roots relation already determines it. A common mistake is to write 6 + p = p; the correct coefficient rule is 6 + p = -p, including the minus sign. Exam cue: whenever the equation is x² + px + q, immediately write root sum = -p.

By Vieta's formulas for \(x^2-20x+96=0\), we have \(\alpha+\beta=20\) and \(\alpha\beta=96\). Thus
\((\alpha-8)(\beta-8)=\alpha\beta-8(\alpha+\beta)+64=96-8\times20+64=96-160+64=0.\)
Notes on distractors: 96 is a common mistake from taking \(\alpha\beta\) directly; -64 can arise from a sign error when handling the constant term. Exam tip: Always compute sum and product of roots first (Vieta) and then substitute into the expanded expression to avoid algebraic slips.

Area of the triangle is given by \(\frac{1}{2}\times\text{base}\times\text{height}\). So \(\frac{1}{2}(x+5)(x+9)=95\). Multiplying both sides by 2 gives \((x+5)(x+9)=190\). Expanding yields \(x^2+14x+45=190\), which simplifies to \(x^2+14x-145=0\); hence option A is correct. Option B is the common mistake of setting \((x+5)(x+9)=95\) (forgetting the factor 1/2). Options C and D have incorrect constant terms. Exam tip: write the area formula first and multiply by 2 immediately to avoid sign/constant errors when forming the quadratic.

For a quadratic \(ax^2+bx+c=0\), the sum of roots is \(\alpha+\beta=-\dfrac{b}{a}\) and the product is \(\alpha\beta=\dfrac{c}{a}\). In this equation \(a=1,\; b=-19,\; c=84\), so \(\alpha+\beta=-\dfrac{-19}{1}=19\). Therefore \((\alpha+\beta)^2=19^2=361\). Option B (84) is the product \(\alpha\beta\), not the square of the sum. Options C (256) and D (324) are \(16^2\) and \(18^2\) respectively and are incorrect because the sum is 19, not 16 or 18. Exam tip: identify \(a,b,c\) quickly and use \(\alpha+\beta=-b/a\); watch the sign of \(b\).

For no real roots the discriminant \(D=b^2-4ac\) must be negative. Here \(a=1,\; b=14,\; c=k\), so \(D=14^2-4\cdot1\cdot k=196-4k\). Requiring \(D<0\) gives \(196-4k<0\Rightarrow k>49\). Thus \(k>49\) is correct. The closest distractor \(k=49\) yields \(D=0\), giving one repeated real root, not “no real roots.” Exam tip: compute \(D\) first and use its sign to decide the nature of roots (\(D<0,=0,>0\)).

Complete the square: \(x^2-2ax+a^2-49=(x-a)^2-49=0\). So \((x-a)^2=49\) and the roots are \(a+7\) and \(a-7\). Their difference is \((a+7)-(a-7)=14\). Option B (7) is incorrect — it is half the actual difference. Option C (2a) is unrelated to the difference here (it would be a mistaken use of coefficients), and option D (a+7) is just one root, not the difference. Exam tip: try completing the square or factor as \((x-a-7)(x-a+7)=0\) to read off roots quickly.

The direct answer is A: \(x^2+x-156=0\), with the restrictions \(x\ne2,-3\). Multiply by the common denominator \(6(x-2)(x+3)\): \(6(x+3)^2+6(x-2)^2=13(x-2)(x+3)\). Expand the left side: \(6(x^2+6x+9+x^2-4x+4)=12x^2+12x+78\). The right side is \(13(x^2+x-6)=13x^2+13x-78\). Subtracting gives \(x^2+x-156=0\). Option A is correct. Option B has the wrong sign of the linear term. Option C has the wrong sign of the constant term. Option D incorrectly keeps a coefficient 2 on \(x^2\). The denominator restrictions must be remembered because multiplying can otherwise introduce invalid values.

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha+\beta=\frac{11}{3}) and (\alpha\beta=2), so the value is (\frac{85}{18}).

The sum of roots is (2k+5) and the product is (k^2+5k+6). ((k+2)+(k+3)=2k+5) and ((k+2)(k+3)=k^2+5k+6) are correct.

By Vieta's relations for \(x^2+px+q=0\), the sum of roots equals \(-p\). The given roots sum to \((p+2)+(q-2)=p+q=8\), so \(-p=8\) and hence \(p=-8\). Note: checking the product condition \((p+2)(q-2)=q\) with \(p=-8\) (so \(q=16\)) gives \(-84\neq16\), so the full set of data is inconsistent; nevertheless the value of \(p\) is determined by the sum relation asked in the question. Exam tip: check both sum and product; a common mistake is to miss the negative sign and pick \(8\).