Introduction to Quadratic Equations

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.

In the first option, the sum is (-\frac{b}{a}=-5) and the product is (\frac{c}{a}=-24). So both are negative.

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.

For equal roots, (p^2-144=0) gives (p=\pm12). For equal positive roots, (-\frac{p}{2}>0), so (p=-12).

Putting (x=-2), the left side becomes (4k-10+6=4k-4). For it not to be a root, (4k-4\neq0), so (k\neq1).

The monic equation is (x^2-(\text{sum})x+\text{product}=0). Using sum (0) and product (-144) gives (x^2-144=0).

The product is (9) and the sum is \(-\frac{k}{5}\). From \(9=-3\left(-\frac{k}{5}\right)\), we get \(k=15\).

The sum of roots is (6) and the product is (-16). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=216+288=504).

For the quadratic equation \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\). Here, the sum is \(-5+11=6\), so \(-b=6\), giving \(b=-6\). Their product is \((-5)(11)=-55\), so \(c=-55\). Therefore, \(b+c=-6-55=-61\). Exam tip: In a monic quadratic, use the root relations sum \(=-b\) and product \(=c\) directly.

Substituting a = 3 gives a − 3 = 0 and a² − 9 = 9 − 9 = 0. Therefore, the equation reduces to 11 = 0, which is a false and contradictory statement, so it has no solution. Exam tip: When the coefficients of the variable terms become zero, check whether the remaining constant statement is true or false; a false statement has no solution.

Expanding gives left side (2x^2-2x+61) and right side (x^2+4x+4). Subtracting gives (x^2-6x+57=0).

Since \(x=-2\) is a root, substituting it into the equation gives \((-2)^2+(5k-2)(-2)+4k=0\). Therefore, \(4-10k+4+4k=0\), so \(8-6k=0\) and \(k=\frac{4}{3}\). Hence, option D is correct. Exam tip: When a root is given, substitute it directly into the quadratic equation and solve for the parameter.

By Vieta’s formulas for \(x^2-8x+5=0\), the sum of the roots is \(\alpha+\beta=8\) and their product is \(\alpha\beta=5\). Therefore, \(\alpha+\beta+4\alpha\beta=8+4(5)=28\). Option 13 results from adding the product only once, so it is incorrect. Exam tip: for \(ax^2+bx+c=0\), the sum of roots is \(-b/a\) and the product is \(c/a\).