Introduction to Quadratic Equations

Direct answer: Option A, p=3. For a quadratic equation \\(ax^2+bx+c=0\\), the product of roots is \\(c/a\\). Here a=p and c=9, so the product is \\(9/p\\). The question says this product equals 3: \\(9/p=3\\). Multiplying by p gives \\(9=3p\\), and dividing by 3 gives \\(p=3\\). Option A is correct. Option B, 6, would give product \\(9/6\\), not 3. Option C, 9, would give product 1. Option D, -3, would give product -3, not the stated positive value. Also p cannot be zero because then the equation would not be quadratic and division by p would be invalid. Remember: product of roots is constant term divided by the coefficient of the squared term.

Check each equation by substituting x = -2.
A: (-2)^2 + 2(-2) = 4 - 4 = 0 → x = -2 is a root.
B: (-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0 → x = -2 is a root.
C: 2(-2)^2 + 3(-2) - 2 = 8 - 6 - 2 = 0 → x = -2 is a root (this is the trickiest distractor because coefficients look nontrivial).
D: (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12 ≠ 0 → x = -2 is not a root.
Therefore option D is correct. Exam tip: For testing a single candidate root, direct substitution is quicker and less error-prone than factoring.

Factorise: \(x^2+2x-35=(x+7)(x-5)=0\). Hence the roots are \(x=-7\) and \(x=5\), i.e. the pair \((5,-7)\). Check: sum \(5+(-7)=-2=-b/a\) and product \(5\times(-7)=-35=c/a\). Option B \((7,-5)\) has the correct product but wrong sum (+2), so it is incorrect. Exam tip: verify roots quickly using sum and product of roots or factorisation before finalising the answer.

If the roots are 6 and 6, their sum is 6+6=12 and product is 6×6=36. For the quadratic \(x^2+rx+36=0\), the sum of roots equals \(-r\) and the product equals 36. Hence \(-r=12\) gives \(r=-12\). Option A (12) mistakes the sign. Options C and D are the root values themselves, not the coefficient asked for. Exam tip: Always apply sum = -b/a and product = c/a, and check signs carefully.

Expand and combine like terms: \((x-3)^2=x^2-6x+9\) and \(2(x-3)=2x-6\). Adding and subtracting 8 gives \(x^2-6x+9+2x-6-8=x^2-4x-5\). Therefore \(x^2-4x-5=0\) is correct. Closest distractor D has the same linear term but wrong constant sign (+5 instead of -5). Exam tip: Always expand \((x-a)^2\) as \(x^2-2ax+a^2\) and then combine like terms carefully.

First expand the left side: \((x+2)(x-7)=x^2-7x+2x-14=x^2-5x-14\). To put in standard form \(ax^2+bx+c=0\), subtract \(3x\) from both sides: \(x^2-5x-14-3x=0\Rightarrow x^2-8x-14=0\). Option B (\(x^2-5x-14=0\)) is the expansion before moving \(3x\), so it is the closest distractor. Options C and D result from sign or arithmetic mistakes. Exam tip: Expand first, then bring all terms to one side and order by descending powers of \(x\).

Substitute \(x=5\): the left-hand side becomes \(5^2-12\cdot5+35=25-60+35=0\). Thus the value is 0 and \(x=5\) is a root. Closest distractors are incorrect because they are either the substituted value (5) or give nonzero results when substituted; always verify by direct substitution and careful arithmetic. Exam tip: perform the arithmetic step-by-step to avoid sign mistakes.

Direct answer: Option B, k=49. Equal roots occur when the discriminant is zero. For \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here a=1, b=-14, and c=k, so \\(D=(-14)^2-4(1)(k)=196-4k\\). Set it equal to zero: \\(196-4k=0\\), hence \\(4k=196\\) and \\(k=49\\). Option B is correct. Option A, 14, does not make the discriminant zero. Option C, 98, gives a negative discriminant, so the roots are not equal real roots. Option D, 196, also fails the zero-discriminant condition. The value 49 also agrees with the repeated-root form \\(x^2-14x+49=(x-7)^2\\).

For equal roots, (k^2=196), and the equal root is (-\frac{k}{2}). For a positive root, (k=-14) is correct.

In (3x^2-27=0), the (x^2) term is present and the (x) term is absent. An equation can be quadratic even without the (x) term.

Dividing every term by (6) gives (x^2-3x+2=0). Dividing by a common nonzero factor does not change the roots.

We need two integers m and n such that m·n = 42 and m + n = 13. Checking 6 and 7 gives 6·7 = 42 and 6+7 = 13, so the quadratic factors as (x-6)(x-7), which expands to x^2 - 13x + 42. A close distractor is 3 and 14: their product is 42 but 3+14 = 17, not 13, so they do not produce the correct middle term. Exam tip: list factor pairs of the constant term and check which pair sums to the coefficient of x (taking sign into account).

For a quadratic \(ax^2+bx+c=0\), the sum of roots = \(-b/a\) and product = \(c/a\). Here \(a=1\) and \(b=s\), so the sum of roots is \(-s\). The given roots sum to \(-3)+(-6)=-9\), therefore \(-s=-9\) which gives \(s=9\). Common wrong answers: \(-9\) arises from forgetting the negative sign, \(18\) confuses sum with product (since product = 18), and \(-6\) mistakes a single root for the coefficient. Exam tip: identify \(a,b,c\) first and apply "sum = -b/a, product = c/a."

For a square with side x the area is \(x^2\). The statement says the area is 15 more than 8 times the side, so \(x^2=8x+15\). Bringing all terms to one side gives \(x^2-8x-15=0\), so option A is correct. Distractors fail for clear reasons: C swaps the coefficients (places 15 with 8) and thus misrepresents the given relation; B has wrong signs (+) so it does not match the sentence. Exam tip: Always translate the English sentence into a mathematical equality (area = ...) before rearranging to standard quadratic form.

Use the identity \((x+a)^2 = x^2 + 2ax + a^2\). Here the middle coefficient 12 gives \(2a=12\) so \(a=6\), and \(a^2=36\), hence \(x^2+12x+36=(x+6)^2\). Option B, \((x-6)^2\), expands to \(x^2-12x+36\) so its middle term has the opposite sign and is incorrect. Exam tip: half the middle coefficient and square it — if that equals the constant term and signs match, it's a perfect square trinomial.

Here ((2x-1)^2=4x^2-4x+1), and bringing all terms to one side gives (4x^2-7x-4=0). Apply the square identity and signs carefully.

Substitute \(x=-4\) into the equation: \((-4)^2 + t(-4) - 8 = 0\), which gives \(16 - 4t - 8 = 0\). Thus \(8 - 4t = 0\) and \(t = 2\). The common wrong choice \(-2\) typically arises from a sign error (writing \(+4t\) instead of \(-4t\)). Exam tip: directly substitute the root and carefully handle squares and signs when simplifying.

Expanding \((4x-1)(x+3)\) gives \(4x^2+12x-x-3\). Combining like terms yields \(4x^2+11x-3\), so the standard form is \(4x^2+11x-3=0\). Option B has the wrong sign for the middle term (should be +11x). Options C and D result from incorrect multiplication or sign errors. Exam tip: multiply each term carefully and then combine like terms to avoid sign mistakes.

The direct answer is B: \(p\ne5\). A quadratic equation must contain a nonzero coefficient of \(x^2\). In this equation that coefficient is \(5-p\). Therefore we require \(5-p\ne0\). Solving this condition gives \(p\ne5\). Option A is wrong because if \(p=5\), the \(x^2\) term disappears and the equation becomes \(2x+9=0\), which is linear, not quadratic. Option B is correct because every value except 5 keeps the leading coefficient nonzero. Option C, \(p=0\), is unnecessary; zero is allowed and gives coefficient 5. Option D, \(p\ne-5\), is not the relevant condition; \(p=-5\) gives coefficient 10 and is perfectly valid. Exam cue: check the coefficient of the highest required power.

The coefficients are \(a=6,\; b=-13,\; c=5\). So \(2a+b-c=2\times6+(-13)-5=12-13-5=-6\). Option B (4) is a common mistake if the sign of \(c\) is handled incorrectly (adding instead of subtracting). Exam tip: always write down the values of \(a,b,c\) first and track plus/minus signs carefully when substituting.

Rewriting the polynomial in descending powers gives \(x^2-7x+12=0\). The coefficient of \(x\) is therefore \(b=-7\). Option C (7) is a common mistake — it ignores the negative sign of the \(x\)-term. Exam tip: arrange terms in order of descending powers (\(x^2, x,\) constant) to read off coefficients directly.

Substitute \(x=5\): \(5^2-6\cdot5+5=25-30+5=0\). Since substitution yields zero, \(x=5\) is a root. Option C is incorrect because the quadratic factors as \((x-1)(x-5)=0\), so both \(x=1\) and \(x=5\) are roots. Exam tip: Verify a proposed root by direct substitution or factor the quadratic to find all roots.

Expand the left side: \((x-6)^2 = x^2 - 12x + 36\). Move the right side \(4x+5\) to the left and combine like terms: \(x^2-12x+36-(4x+5)=x^2-16x+31\). Thus the standard form is \(x^2-16x+31=0\). The closest distractor C (\(x^2-12x+41=0\)) keeps the middle term but has an incorrect constant — a typical arithmetic slip when subtracting the constant. Exam tip: always bring all terms to one side and simplify carefully, checking signs and arithmetic.

Use the discriminant formula \(D=b^2-4ac\). Here \(a=25,\; b=-20,\; c=4\), so \(D=(-20)^2-4\cdot25\cdot4=400-400=0\). \(D=0\) means the quadratic has two real and equal roots (a repeated root). The distractor 400 is incorrect because it is just \(b^2\) without subtracting \(4ac\). Option 100 arises from a common arithmetic mistake in computing \(4ac\), and \(-20\) comes from sign or substitution errors. Exam tip: check for a perfect square trinomial — here \(25x^2-20x+4=(5x-2)^2\), which immediately shows \(D=0\).

Compute the discriminant \(\Delta=b^2-4ac\). Here \(a=1,\; b=14,\; c=49\) so \(\Delta=14^2-4\cdot1\cdot49=196-196=0\). When \(\Delta=0\) a quadratic has two equal (repeated) real roots; using \(x=-\dfrac{b}{2a}\) gives \(x=-7\) as the repeated root. Why other options are wrong: A would require \(\Delta>0\), C would require \(\Delta<0\), and D is impossible for a quadratic (at most two roots). Exam tip: Always evaluate \(\Delta\) first — it directly tells you whether roots are distinct, equal, or complex.