Introduction to Quadratic Equations

Here (\alpha+\beta=10) and (\alpha\beta=q). Using (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta), (58=100-2q), so (q=21).

The area of a rectangle is length × breadth, so \\((2x+1)(x-4)=45\\). Expanding the product gives \\(2x^2-8x+x-4=2x^2-7x-4\\). Therefore, \\(2x^2-7x-4=45\\), which simplifies to \\(2x^2-7x-49=0\\). Hence, option A is correct. Exam tip: For area-based questions, first form the product equation and then bring all terms to one side to make it equal to zero.

Here ((3x-2)(2x+5)=6x^2+11x-10) and (4(x+1)=4x+4). Bringing all terms to one side gives (6x^2+7x-14=0).

For the equation to be quadratic, the coefficient of (x^2) must not be (0). Here (r^2-49\neq0), so (r\neq\pm7).

Expanding the brackets gives 2x² − 4x + 2 + 3x² + 12x + 12 = 65. Combining like terms, 5x² + 8x + 14 = 65, and moving 65 to the left gives 5x² + 8x − 51 = 0. Therefore, option A is correct. Option B has an incorrect sign for the linear term. Exam tip: Expand each squared binomial carefully before rearranging the equation into ax² + bx + c = 0.

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.

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta). Here ((-2)^2-2\cdot\frac{3}{4}=\frac{5}{2}).

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.

From (2r+5r=21), we get (r=3), so the roots are (6) and (15). The constant term is the product of roots (90).

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.