Introduction to Quadratic Equations

Correct answer: A. x² − 12x + 36 = 0

In option A, x² − 12x + 36 = (x − 6)², so the repeated root is x = 6, which is positive. Option B has equal roots of −6, option C has two distinct roots, 6 and −6, and option D has no real roots. Exam tip: equal roots require a discriminant of zero, and the repeated root must be positive.

Correct answer: A. \(a=5\)

Expand \((x+a)^2\): \((x+a)^2=x^2+2ax+a^2\). Match with \(x^2+10x+25\): the coefficient of \(x\) gives \(2a=10\Rightarrow a=5\). (Then \(a^2=25\) also matches.)

Correct answer: A. Two equal real roots

The equation can be written as \(4x^2-4kx+k^2=(2x-k)^2=0\). Hence \(2x-k=0\), giving the repeated root \(x=\frac{k}{2}\). Therefore, the roots are equal and real. Exam tip: use the discriminant \(D=b^2-4ac\); here \(D=0\), so the roots are equal. Thus option B is incorrect because the roots are not distinct.

Correct answer: A. It is not possible

If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.

Correct answer: A. (6)

For equal roots, (p^2-36=0) gives (p=\pm6). For equal negative roots, (-\frac{p}{2}<0), so (p=6).

Correct answer: A. (x^2-49=0)

The monic equation is (x^2-(\text{sum})x+\text{product}=0). Using sum (0) and product (-49) gives (x^2-49=0).

Correct answer: A. (6)

The direct answer is A: \(k=6\). For \(2x^2+kx+18=0\), the sum of roots is \(-k/2\), and the product is \(18/2=9\). The statement says product equals -3 times the sum, so \(9=-3(-k/2)=3k/2\). Multiplying by 2 gives \(18=3k\), hence \(k=6\). Option A is correct. Option B, \(-6\), makes the sum 3, so -3 times it is -9, not 9. Option C, 12, gives product relation 18, not 9. Option D, -12, gives the wrong sign and magnitude. The key is to use Vieta’s formulas with signs carefully. Memory cue: sum is \(-b/a\), product is \(c/a\).

Correct answer: A. (26)

The sum of roots is (2) and the product is (-3). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=8+18=26).

Correct answer: A. -19

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 \\(-2+7=5\\), so \\(-b=5\\), giving \\(b=-5\\). The product is \\((-2)\\times7=-14\\), so \\(c=-14\\). Therefore, \\(b+c=-5-14=-19\\). Exam tip: For \\(ax^2+bx+c=0\\), use the root sum \\(-b/a\\) and root product \\(c/a\\).

Correct answer: A. \(x^2-6x=0\)

In option A, \(x^2-6x=x(x-6)\), so the roots are \(x=0\) and \(x=6\). Since \(6\) is positive, this option is correct. Option B has roots \(0\) and \(-6\), so its other root is negative; option C has no real roots, and option D does not have zero as a root. A quick exam check is that an equation with \(x=0\) as a root must have a zero constant term.

Correct answer: C. Contradictory statement

Substituting \\(a=-1\\) gives \\(a+1=0\\) and \\(a^2-1=1-1=0\\). Hence the equation becomes \\(0x^2+0x+2=0\\), or \\(2=0\\). This is a false, contradictory statement and has no solution. It is not a linear equation because the coefficient of \\(x\\) is also zero. In parameter-based equations, first check the coefficients of \\(x^2\\), \\(x\\), and the constant term separately.

Correct answer: B. (x^2-2x-4=0)

Expanding gives left side (2x^2-6x+5) and right side (x^2-6x+9). Subtracting gives (x^2-4=0).

Correct answer: A. 0

A root must make the left-hand side of the equation equal to zero. Substituting \(x=1\) gives \(1+(2k-1)+k=0\), so \(3k=0\) and hence \(k=0\). The values \(\pm\frac{1}{3}\) result from incorrect simplification. Exam tip: substitute the given root carefully and collect the parameter terms before solving.

Correct answer: A. 5

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\). Here, \(a=1, b=-4, c=1\), so \(\alpha+\beta=4\) and \(\alpha\beta=1\). Therefore, \(\alpha+\beta+\alpha\beta=4+1=5\). Exam tip: Use the sum and product of roots directly instead of solving the quadratic equation.

Correct answer: A. \(q=2\)

For \(x^2+px+q=0\) with roots \(\alpha\) and \(\beta\), we have \(q=\alpha\beta\). Here \(\alpha=1\) and \(\beta=2\), so \(q=1\cdot 2=2\).

Correct answer: B. \(x(x-3)=4\)

Bringing \(x(x-3)=4\) to one side gives \(x^2-3x-4=0\), where the coefficient of \(x^2\) is 1. Hence it is quadratic. Option D is an identity because both sides are identical. Exam tip: check that the highest power is 2.

Correct answer: A. (2x^2+3x-5=0)

Direct answer: Option A, \\(2x^2+3x-5=0\\). For \\(ax^2+bx+c=0\\), the product of roots is \\(c/a\\). We only need the sign. Option A has a=2 and c=-5, so the product is \\(-5/2\\), which is negative. Option B has c=5 and a=2, giving \\(5/2>0\\), so it is not negative. Option C also has a=2 and c=5, so its product is again positive; the changed sign of the x term does not affect the product. Option D has c=0, so its product is 0, not negative. Thus A is the only correct choice. Do not confuse the sign of b with the sign of the product: the product depends on c/a.

Correct answer: A. \(x^2+4x-57=0\)

The area of a right triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+1)(x+3)=30\), giving \((x+1)(x+3)=60\). Expanding, \(x^2+4x+3=60\), so the quadratic equation is \(x^2+4x-57=0\). Option D results from mishandling the factor \(\frac{1}{2}\). Exam tip: For triangle-area problems, include the factor \(\frac{1}{2}\) before expanding the algebraic expression.

Correct answer: A. 4

By Vieta’s formula, the sum of the roots of \(ax^2+bx+c=0\) is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=1\) and \(b=2\), so \(\alpha+\beta=-2\). Therefore, \((\alpha+\beta)^2=(-2)^2=4\). Exam tip: square the complete sum, including its sign; the negative sign disappears only after squaring.

Correct answer: A. \(k>4\)

A quadratic equation has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=4, c=k\), so \(D=4^2-4(1)(k)=16-4k\). Thus, \(16-4k<0\), which gives \(k>4\). At \(k=4\), the discriminant is zero and the equation has one repeated real root, so that option is incorrect. Exam tip: for ‘no real roots,’ always use the condition \(D<0\).

Correct answer: A. 6

The equation can be rewritten as \((x-a)^2-9=0\). Thus, \((x-a)^2=9\), giving the roots \(x=a+3\) and \(x=a-3\). The difference between the larger and smaller roots is \((a+3)-(a-3)=6\). Option D represents only one root, while option C is not generally the difference between the roots. Exam tip: For this form, complete the square and identify the two roots directly.

Correct answer: A. (6x^2-28x-4=0)

Here ((6x+1)(x-4)=6x^2-23x-4), and subtracting (5x) gives (6x^2-28x-4=0). First expand and then bring all terms to one side.

Correct answer: A. \\(5x^2-2x=0\\)

Expand the squares: \\( (2x-3)^2=4x^2-12x+9 \\) and \\( (x+5)^2=x^2+10x+25 \\). This gives \\(5x^2-2x+34=34\\); subtracting 34 from both sides yields \\(5x^2-2x=0\\). This matches the standard quadratic form \\(ax^2+bx+c=0\\), with \\(c=0\\). Exam tip: after expansion, bring all terms to one side and equate the result to zero.

Correct answer: A. (2x^2-3x-22=0)

Multiplying the whole equation by (6) gives (2x^2+2-3x+6=30). Therefore the standard form is (2x^2-3x-22=0).

Correct answer: A. 5

A root makes the equation equal to zero. Substitute x = −4: 2(−4)² + p(−4) − 12 = 0, so 32 − 4p − 12 = 0. Thus 20 − 4p = 0, giving p = 5.