Introduction to Quadratic Equations

The direct answer is A: \(k=6\). For \(3x^2+kx+12=0\), the product of roots is \(12/3=4\), while their sum is \(-k/3\). The condition says \(4=-2\) times the sum, so \(4=-2(-k/3)=2k/3\). Multiplying by 3 gives \(12=2k\), hence \(k=6\). Option A works exactly. Option B, \(-6\), makes the sum 2 and -2 times it -4, not 4. Option C, 12, gives the right side 8. Option D, -12, gives -8. The signs come from the minus sign in the sum formula. Exam cue: write \(S=-b/a\) and \(P=c/a\) before substituting.

The sum of roots is (3) and the product is (-10). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=27+90=117).

For the monic quadratic equation \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\). Here, the sum is \(-3+8=5\), so \(-b=5\), giving \(b=-5\). The product is \((-3)(8)=-24\), so \(c=-24\). Therefore, \(b+c=-5-24=-29\). Exam tip: For \(x^2+bx+c=0\), use \(b=-(\text{sum of roots})\) and \(c=\text{product of roots}\).

In option A, \(x^2+7x=x(x+7)\). Hence, the roots are \(x=0\) and \(x=-7\), so the second root is negative. Option B has roots \(0\) and \(7\), and therefore does not satisfy the condition. Exam tip: For an equation of the form \(x^2+bx=0\), the roots are \(0\) and \(-b\).

Substituting \(a=2\) gives \((2-2)x^2+(2^2-4)x+5=0\), or \(0x^2+0x+5=0\), which reduces to \(5=0\). This statement is false and is satisfied by no value of \(x\), so it is a contradictory statement. It is not linear because the coefficient of \(x\) is also zero, and it is not always true. Exam tip: After substituting a parameter value, check the highest non-zero power and whether the resulting constant statement is true or false.

Expanding gives left side (2x^2-6x+29) and right side (x^2+2x+1). Subtracting gives (x^2-8x+28=0).

Since \(x=-2\) is a root, substitute it into the equation: \((-2)^2+(3k+1)(-2)+2k=0\). This gives \(4-6k-2+2k=0\), so \(2-4k=0\) and hence \(k=\frac{1}{2}\). The nearby distractor \(k=-\frac{1}{2}\) does not make the equation equal to zero. Exam tip: when a root is given, substitute it directly into the polynomial equation.

For a quadratic equation \\(ax^2+bx+c=0\\), the sum of the roots is \\( -b/a \\) and their product is \\(c/a\\). Here, \\(a=1,b=-5,c=2\\), so \\(\\alpha+\\beta=5\\) and \\(\\alpha\\beta=2\\). Therefore, \\(\\alpha+\\beta+2\\alpha\\beta=5+2(2)=9\\). Exam tip: use the sum and product of roots directly instead of solving for the individual roots.

For the monic quadratic equation (x^2+px+q=0), the product of the roots equals the constant term (q). Since the roots are (2) and (p), (q=2\times p=2p). Also, their sum gives (2+p=-p), so (p=-1) and (q=-2), confirming the relation. Exam tip: In (x^2+ax+b=0), the sum of roots is (-a) and their product is (b).

By Vieta’s formulas, \alpha+\beta=9 and \alpha\beta=20. Therefore, (\alpha-4)(\beta-4)=\alpha\beta-4(\alpha+\beta)+16=20-4(9)+16=0. Hence, option A is correct. Choosing 4 or -16 usually results from omitting the middle term -4(\alpha+\beta) during expansion. Exam tip: for (ax^2+bx+c=0), the sum of roots is -b/a and their product is c/a.

The area of a triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Therefore, \(\frac{1}{2}(x+2)(x+6)=40\), giving \((x+2)(x+6)=80\). Expanding, \(x^2+8x+12=80\), so the required equation is \(x^2+8x-68=0\). Option B incorrectly uses 40 instead of 80 after removing the \(\frac{1}{2}\) factor. Exam tip: write the area formula first, then expand and collect all terms on one side.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=1\) and \(b=-9\), so \(\alpha+\beta=9\). Therefore, \((\alpha+eta)^2=9^2=81\). The value 14 is the product \(\alpha\beta\), not the sum of the roots. Exam tip: remember that the sum of roots is \(-b/a\), while their product is \(c/a\).

The direct answer is option A: -13. For a monic quadratic x² + bx + c = 0 with roots α and β, the sum of roots is α + β = -b and the product is αβ = c. In x² - 4x - 21 = 0, b = -4 and c = -21. Therefore α + β = -(-4) = 4 and αβ = -21. The required expression is αβ + 2α + 2β. Group the last two terms: αβ + 2(α + β). Substitute the known values: -21 + 2(4) = -21 + 8 = -13. Thus option A is correct. Option B, -21, is only the product αβ and ignores the additional terms. Option C, 13, has the wrong sign and does not follow from the calculation. Option D, 29, is obtained by an incorrect operation, such as adding quantities with the wrong sign. The useful lesson is to identify the sum and product first rather than solving the quadratic roots individually. Memory cue: for x² + bx + c, remember “sum opposite of b, product c.”

A quadratic equation has no real roots only when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=8, c=k\), so \(D=8^2-4(1)(k)=64-4k\). Thus, \(64-4k<0\), which gives \(k>16\). Note that \(k=16\) makes the discriminant zero and produces one repeated real root, so it is not correct.

The equation can be rewritten as \((x-a)^2-16=0\). Thus, \((x-a)^2=16\), giving the roots \(x=a+4\) and \(x=a-4\). Their difference is \((a+4)-(a-4)=8\). The value 4 represents the distance of each root from \(a\), not the distance between the two roots. In the exam, completing the square is a quick method for such questions.

The sum of roots is (3+p), and in the equation the sum is (-p). Thus (3+p=-p), giving (p=-\frac{3}{2}).

Expanding the two products gives \((x-2)(x+3)=x^2+x-6\) and \((2x-1)(x-4)=2x^2-9x+4\). Combining like terms, \(x^2+2x^2=3x^2\), \(x-9x=-8x\), and \(-6+4=-2\). Hence the standard form is \(3x^2-8x-2=0\). Option B has the wrong sign for the linear term. Exam tip: expand each bracket carefully, then combine the quadratic, linear, and constant terms separately.

The original equation requires x ≠ 1 and x ≠ −2 because these values make a denominator zero. Multiply both sides by 2(x − 1)(x + 2), which is valid under those restrictions. This gives 2(x + 2)² + 2(x − 1)² = 5(x − 1)(x + 2). Expanding the left side: 2(x² + 4x + 4) + 2(x² − 2x + 1) = 4x² + 4x + 10. Expanding the right side gives 5(x² + x − 2) = 5x² + 5x − 10. Moving the left side to the right yields 0 = x² + x − 20, or x² + x − 20 = 0. Thus option A is correct; the other options have an incorrect sign, constant, or leading coefficient.