Introduction to Quadratic Equations

Multiplying the whole equation by (12) gives (9x^2+6-4x+20=72). Therefore the standard form is (9x^2-4x-46=0).

Since \(x=-5\) is a root, substitute it directly into the equation: \(3(-5)^2+p(-5)-20=0\). Thus, \(75-5p-20=0\), so \(55-5p=0\) and \(p=11\). The distractor \(-11\) results from mishandling the negative sign in \(p(-5)\). Exam tip: substitute the given root carefully and simplify before solving for the parameter.

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-13) gives (x^2+13x+42=0).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here \(a=1, b=-2m, c=49\), so \(D=(-2m)^2-4(1)(49)=4m^2-196=0\). Thus, \(m^2=49\), giving \(m=\pm7\). Choosing only \(m=7\) or only \(m=-7\) omits one valid value. Exam tip: For equal roots of a quadratic equation, set the discriminant directly equal to zero.

A quadratic equation has real roots when its discriminant satisfies \(D\geq 0\). Here, \(a=1, b=2p, c=36\), so \(D=(2p)^2-4(1)(36)=4p^2-144\). Thus, \(4p^2-144\geq 0\), which gives \(p^2\geq 36\) and hence \(p\leq -6\) or \(p\geq 6\). In option B, the discriminant is negative, so the roots are not real. Exam tip: include \(D=0\), because equal roots are also real roots.

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here it is (\frac{\frac{17}{5}}{\frac{6}{5}}=\frac{17}{6}).

By Vieta’s formulas, the sum of the roots is \(s=7+8=15\), and their product is \(p=7\times8=56\). Therefore, \(s+p=15+56=71\), so option A is correct. Exam tip: In an equation of the form \(x^2-sx+p=0\), the sum of the roots is \(s\) and their product is \(p\).

A quadratic equation has real and distinct roots only when its discriminant is positive. Here, \(D=(-16)^2-4(1)(k)=256-4k\). Thus, \(256-4k>0\), which gives \(k<64\). At \(k=64\), the discriminant is zero and the roots are equal, so option B is not correct. Exam tip: For real and distinct roots, always apply the condition \(D>0\).

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. In option A, a = 1, b = 1, and c = 9, so D = 1² − 4 × 1 × 9 = 1 − 36 = −35. For example, option C gives 2² − 4 × 3 × 3 = −32, so it is not correct. Exam tip: the sign of b disappears when calculating b², but the complete value of 4ac must be subtracted.

The direct answer is option A, \(5/2\). Let the roots of \(4x^2+8x+3=0\) be \(\alpha\) and \(\beta\). Using the root relations, \(\alpha+\beta=-8/4=-2\) and \(\alpha\beta=3/4\). The identity for the sum of squares is \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\). Substitution gives \((-2)^2-2(3/4)=4-3/2=5/2\). Hence option A is correct. Option B, 4, is only the square of the sum and forgets the subtraction of \(2\alpha\beta\). Option C, \(3/4\), is the product of the roots, not the sum of their squares. Option D, \(13/4\), results from an incorrect operation and does not match the identity. The roots themselves need not be solved. Memory cue: square-sum = square of sum − twice product.

By Vieta’s formulas, \(\alpha+\beta=14\) and \(\alpha\beta=48\). Therefore, \((\alpha-6)(\beta-6)=\alpha\beta-6(\alpha+\beta)+36=48-6(14)+36=0\). Hence, option A is correct. Option C is only the value of \(\alpha\beta\), not of the complete expression. Exam tip: In such questions, use the sum and product of the roots instead of finding the roots separately.

Expand the two products: (x+4)(x+7)=x²+11x+28 and (x−3)(x+6)=x²+3x−18. Their sum is 2x²+14x+10, so the equation becomes 2x²+14x+10=88. Bringing 88 to the left gives 2x²+14x−78=0, so option A is correct. Option B stops before subtracting 88. Exam tip: To write a quadratic equation in standard form, bring all terms to one side so that the other side is 0.

The direct answer is option A: 90. A monic quadratic has leading coefficient 1, so if its roots are r₁ and r₂, its equation is x² - (r₁ + r₂)x + r₁r₂ = 0. Therefore the constant term equals the product of the roots. The roots here are 2r and 5r, and their sum is 21. Thus 2r + 5r = 21, so 7r = 21 and r = 3. The roots are therefore 2(3) = 6 and 5(3) = 15. Their product is 6 × 15 = 90, so the constant term is 90. Option A is correct. Option B, 21, is the sum of the roots, not their product. Option C, 45, does not equal the product 6 × 15 and may result from an incomplete calculation. Option D, 63, is another incorrect product or arithmetic value; it is not the constant term. The word “monic” is important: because the leading coefficient is 1, the constant term directly equals the root product. Memory cue: first find r from the sum, then multiply the two roots.

Here, \(a=1\), \(b=-10\), and \(c=29\). The discriminant is \(D=b^2-4ac=(-10)^2-4(1)(29)=100-116=-16\). Since \(D<0\), the equation has no real roots; its roots are complex. Option B would be correct only if \(D=0\), which is not the case here. Exam tip: For a quadratic equation, \(D<0\) always indicates that there are no real roots.

Expanding \(x(x-3)=4\) gives \(x^2-3x-4=0\), whose highest power of \(x\) is 2, so it is quadratic. Option B is an identity because both sides are identical. Exam tip: bring all terms to one side, then check the highest power.

Factoring gives \(5x^2-22x+24=(5x-12)(x-2)\). Therefore, the roots are \(\frac{12}{5}\) and \(2\). The difference between the larger and smaller roots is \(\frac{12}{5}-2=\frac{2}{5}\). The values \(\frac{12}{5}\) and \(2\) are the roots themselves, not their difference. In an exam, subtract the smaller root from the larger root when the difference between roots is asked.

For the expression to be a perfect square, \(x^2-18x+c\) must equal \((x-9)^2=x^2-18x+81\). Therefore, \(c=81\). Choosing 324 incorrectly amounts to squaring 18 directly; the correct method is to square half the coefficient of \(x\). Exam tip: \(x^2+bx+c\) is a perfect square when \(c=\left(\frac{b}{2}\right)^2\).

By Vieta’s formula, the sum of the roots of \(x^2+bx+100=0\) is \(-b\). Since both roots are \(-10\), their sum is \(-10+(-10)=-20\). Therefore, \(-b=-20\), giving \(b=20\). Exam tip: for \(x^2+bx+c=0\), the sum of the roots is always \(-b\); missing this negative sign leads to option B.

The answer is option A. For a quadratic equation ax^2+bx+c=0, the sum of roots is -b/a and their product is c/a. In option A, a=1, b=5, c=-24, so the sum is -5 and the product is -24; both are negative. Option B has sum 5, although its product is -24, so it fails the first condition. Option C has sum -5, but its product is 24, so it fails the second condition. Option D has sum 5 and product 24, so neither required sign is obtained. Therefore only A works. Remember: the middle coefficient is negated for the sum, but the constant term keeps its sign for the product.

Area of a rectangle = length × breadth, so \((x+9)(x-6)=130\). Expanding gives \(x^2+3x-54=130\), and rearranging produces \(x^2+3x-184=0\). Option B does not correctly account for the \(-54\) term, while options C and D contain errors in expansion. Exam tip: multiply the binomials first, then bring all terms to one side and equate the expression to zero.

If the sum of roots is (17) and product is (72), the equation is (x^2-17x+72=0). Remember the monic form formula.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the value is (\frac{13}{42}).

Expanding gives \\((x+a)^2=x^2+2ax+a^2\\). Comparing the coefficients of \\(x\\) on both sides, \\(2a=-20\\), so \\(a=-10\\). Substituting this value also gives \\(a^2=100\\), confirming the constant term. Exam tip: For an identity, compare coefficients of like powers of the variable.

The equation can be written as 25x² + 40kx + 16k² = (5x + 4k)². Hence, (5x + 4k)² = 0 gives x = −4k/5, which is a repeated root. Equivalently, the discriminant is D = (40k)² − 4(25)(16k²) = 0, so the roots are equal and real. Option B is incorrect because distinct real roots require D > 0. Exam tip: For a quadratic equation, D = 0 indicates two equal real roots.

If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (49), so it is not possible.