Introduction to Quadratic Equations

Correct answer: A. (-12)

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

Correct answer: A. (k\neq1)

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).

Correct answer: A. (x^2-144=0)

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

Correct answer: A. (15)

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

Correct answer: A. (504)

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).

Correct answer: A. -61

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.

Correct answer: A. Contradictory statement

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.

Correct answer: A. (x^2-10x+57=0)

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

Correct answer: D. \(\frac{4}{3}\)

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.

Correct answer: A. 28

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 \(c/a\).

Correct answer: A. -(\frac{5}{2}) / (-\frac{5}{2})

The sum of roots is (5+p), and in the equation the sum is (-p). Thus (5+p=-p), giving (p=-\frac{5}{2}).

Correct answer: A. \(0\)

By Vieta’s formulas, \(\alpha+\beta=16\) and \(\alpha\beta=63\). Therefore, \((\alpha-7)(\beta-7)=\alpha\beta-7(\alpha+\beta)+49=63-7(16)+49=0\). Hence, option A is correct. Exam tip: Use the sum and product of the roots directly instead of finding the roots separately.

Correct answer: A. x² + 12x − 112 = 0

A triangle’s area is half its base times its height, so ½(x + 4)(x + 8) = 72. Multiplying by 2 gives (x + 4)(x + 8) = 144. Expanding and rearranging yields x² + 12x − 112 = 0, which is option A.

Correct answer: A. 225

By Vieta’s formula, the sum of the roots of \(ax^2+bx+c=0\) is \(-\frac{b}{a}\). Here, \(a=1\) and \(b=-15\), so \(\alpha+\beta=-\frac{-15}{1}=15\). Therefore, \((\alpha+\beta)^2=15^2=225\). The value 44 is the product \(\alpha\beta\), not the sum of the roots. Exam tip: Identify the sum and product of roots directly using Vieta’s formulas.

Correct answer: C. \((x+1)^2=x^2+4x\)

In option C, expanding gives \(x^2+2x+1=x^2+4x\). The \(x^2\) terms cancel, leaving \(2x-1=0\), which is linear. Exam tip: simplify fully and then check the highest power of x.

Correct answer: A. 12

The equation can be rewritten as \((x-a)^2-36=0\). Thus, \((x-a)^2=36\), giving the roots \(x=a+6\) and \(x=a-6\). Their difference is \((a+6)-(a-6)=12\). The expression \(a+6\) represents only one root, not the difference. Exam tip: First rewrite such equations in perfect-square form.

Correct answer: A. ( \frac{37}{6} )

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha+\beta=\frac{7}{2}) and (\alpha\beta=\frac{3}{2}), so the value is (\frac{37}{6}).

Correct answer: A. (3)

The sum is (2k+3) and the product is (k^2+3k). Direct checking with the options shows (k=3) gives the required equation pattern.

Correct answer: A. (6)

The sum of roots is (-p), so ((p+1)+(q-1)=p+q=-p). Using (p+q=5), the option consistent with the conditions is (6).

Correct answer: A. (4x^2+11x-13=0)

Here ((4x-3)(x+5)=4x^2+17x-15) and (2(3x-1)=6x-2). Bringing all terms to one side gives (4x^2+11x-13=0).

Correct answer: C. (t\neq \pm8)

For the equation to be quadratic, the coefficient of (x^2) must not be (0). Here (t^2-64\neq0), so (t\neq\pm8).

Correct answer: A. 5x^2-10x-39=0

Expand the squares first: \(3(x+1)^2=3(x^2+2x+1)=3x^2+6x+3\) and \(2(x-4)^2=2(x^2-8x+16)=2x^2-16x+32\). Adding gives \(5x^2-10x+35=74\). Bring 74 to the left and simplify: \(5x^2-10x+35-74=0\), so \(5x^2-10x-39=0\). Option B is the closest distractor because it has the wrong sign on the linear term (+10x instead of -10x). Exam tip: expand carefully, combine like terms, then move the constant from the RHS to the LHS and re-check the arithmetic for the constant term (here \(3+32-74=-39\)).

Correct answer: A. (16x^2+5x-74=0)

Multiplying the whole equation by (20) gives (16x^2-12+5x-10=60). Therefore the standard form is (16x^2+5x-82=0).

Correct answer: A. 9

Substitute \(x=-6\) into \(2x^2+px-18=0\): \(2(-6)^2 + p(-6) -18 = 0\). This gives \(72 - 6p - 18 = 0\) → \(54 - 6p = 0\), so \(p = 54/6 = 9\). The choice \(-9\) typically arises from a sign error (treating \(p(-6)\) as \(+6p\)). Exam tip: always substitute the root and simplify step-by-step, checking signs carefully.

Correct answer: A. (x^2+15x+56=0)

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-15) gives (x^2+15x+56=0).