Introduction to Quadratic Equations

Answer: B, m ≠ 2. A quadratic equation must have a non-zero coefficient of x². In (m − 2)x² + 3x + 1 = 0, that coefficient is m − 2. Therefore require m − 2 ≠ 0. Adding 2 to both sides gives m ≠ 2. If m = 2, the x² term disappears and the equation becomes 3x + 1 = 0, which is linear, not quadratic. If m = 0, the equation becomes −2x² + 3x + 1 = 0, which is still quadratic. If m = 1, it becomes −x² + 3x + 1 = 0, also quadratic. Thus C and D are permitted examples but do not state the complete condition; B includes every allowed value. A is precisely the excluded value. Memory cue: for a parameterised quadratic, set the leading coefficient non-zero, then solve the resulting condition.

Multiplying gives (2x^2+8x-3x-12=0), so (2x^2+5x-12=0). Adding middle terms correctly is important in exams.

The direct answer is option B, \(k \ne -1\). A quadratic equation must contain a non-zero coefficient of \(x^2\). Here that coefficient is \(k+1\). Therefore we require \(k+1 \ne 0\). Subtracting 1 from both sides gives \(k \ne -1\). Option B is correct. Option A, \(k=1\), is only one allowed value, whereas many values other than −1 are allowed; it is not the complete condition. Option C, \(k=-1\), makes \(k+1=0\), so the equation becomes \(-5x+6=0\), which is linear, not quadratic. Option D, \(k\ne1\), wrongly excludes 1 even though when \(k=1\), the coefficient of \(x^2\) is 2 and the equation is quadratic. The key test is the coefficient of the highest required power, not a particular guessed value of k. Memory cue: for a quadratic, coefficient of \(x^2\) must not be zero.

In standard form \(ax^2+bx+c=0\), we have \(a=3,\; b=-2,\; c=7\). Thus \(a+b+c=3+(-2)+7=8\). A common mistake is to drop the negative sign and take \(b=+2\), giving the incorrect sum \(3+2+7=12\) (option B). Exam tip: Always write down coefficients with their signs from the standard form before calculating sums.

Bringing all terms to one side gives (-2x^2+5x-9=0), and multiplying by (-1) gives (2x^2-5x+9=0). Change signs carefully in standard form.

Check a candidate root by substituting it into the expression. Putting \(x=3\) gives \(3^2-4\cdot3+3=9-12+3=0\), so \(x=3\) satisfies the equation and is a root. Factoring gives \(x^2-4x+3=(x-1)(x-3)\), showing the roots are \(x=1\) and \(x=3\); thus option C (only \(x=1\)) is incorrect. Exam tip: verify roots by direct substitution or by factoring the quadratic to list all roots quickly.

Substitute the given root into the equation: \((-2)^2 + p(-2) - 6 = 0\) ⇒ \(4 - 2p - 6 = 0\) ⇒ \(-2 - 2p = 0\) ⇒ \(-2p = 2\) ⇒ \(p = -1\). A common mistake is to drop the negative sign on the \(px\) term, which would incorrectly give \(p=1\) (option A). Options C and D do not satisfy the equation when substituted. Exam tip: always substitute the root and simplify carefully, paying attention to signs.

(x^2=5x-4) can be written as (x^2-5x+4=0), so it is quadratic. In standard form, the right side must be (0).

The direct answer is option A: \\(x^2-10x+9=0\\). Start with \\((x-5)^2=16\\). Use the identity \\((a-b)^2=a^2-2ab+b^2\\): \\((x-5)^2=x^2-10x+25\\). Therefore \\(x^2-10x+25=16\\). Move 16 to the left: \\(x^2-10x+25-16=0\\), so \\(x^2-10x+9=0\\). Option A is correct. Option B, \\(x^2-10x+41=0\\), results from adding 16 instead of subtracting it after bringing all terms to one side. Option C, \\(x^2+10x+9=0\\), has the wrong sign for the middle term; the expression contains \\(x-5\\), so the middle term is negative. Option D, \\(x^2-5x+16=0\\), does not correctly expand the square and also mishandles the constant. Memory cue: when \\(x-5\\) is squared, the middle term is \\(-10x\\), not \\(+10x\\).

Using the standard form \(ax^2+bx+c=0\) and substituting \(a=2, b=-7, c=3\) yields \(2x^2-7x+3=0\). Option A is wrong because it uses +7 instead of −7 (wrong sign for b). Option C swaps coefficients (a and c are interchanged). Option D alters the coefficients arbitrarily. Exam tip: write \(ax^2+bx+c=0\) first and then substitute the given a, b, c to avoid sign errors.

The discriminant tells us the nature of the roots of a quadratic equation \\(ax^2+bx+c=0\\). It is calculated as \\(D=b^2-4ac\\). If \\(D>0\\), the equation has two real and unequal roots. If \\(D=0\\), the roots are real and equal, and if \\(D<0\\), there are no real roots. Here \\(D=25\\), and 25 is positive.

Therefore, the equation has two distinct real roots, so option A is correct. The actual values of the roots are not needed because the question asks only about their nature. A positive discriminant does not reduce the degree of the equation; the equation remains quadratic as long as the coefficient of \\(x^2\\) is nonzero. Thus options B, C, and D do not follow from \\(D=25\\).

The discriminant is defined by \(D=b^2-4ac\). Here \(a=2,\; b=3,\; c=5\). So \(D=3^2-4\cdot2\cdot5=9-40=-31\). Since \(D<0\), the discriminant is negative and the equation has no real roots (it has complex conjugate roots). Option B (zero) is wrong because \(D=0\) would give equal real roots; option A (positive) is wrong because a positive \(D\) would give two distinct real roots. Exam tip: compute the \(4ac\) term first and track signs carefully to avoid arithmetic errors.

The factors are ((x+2)) and ((x-5)), giving (x^2-3x-10=0). Make factors with signs opposite to the roots.

Let the unknown number be \(x\). Its square is \(x^2\). The statement says that the number plus its square equals 42, so the direct equation is \(x+x^2=42\). Rearranging all terms to one side gives \(x^2+x-42=0\), which is option A.

The order of the first two terms does not matter because addition is commutative, but none of the other options represents the stated sum correctly. Option B changes the sign of the number, option C doubles the square instead of adding the number, and option D uses 42 as a coefficient of \(x\). The correct method is to translate each phrase carefully into an algebraic expression before rearranging it. Therefore the supplied answer is correct.

Let the smaller integer be \(x\); the next consecutive integer is \(x+1\). Their product gives \(x(x+1)=56\). Expanding yields \(x^2+x-56=0\), so option A is correct. Distractor B (\(x^2-x-56=0\)) corresponds to integers \(x\) and \(x-1\), not the given consecutive pair. Option C (\(x^2+2x-56=0\)) would arise from \(x\) and \(x+2\), and D is a linear equation unrelated to the product of two integers. Exam tip: after forming the equation, check factor pairs of 56 (e.g., 7 and 8) to verify your result quickly.

Breadth = x, so length = x+4. Area = length × breadth = x(x+4) = 45. Expanding gives x^2+4x-45=0, so option A is correct. Option B is wrong because it assumes length = x−4 (wrong sign). Option C would arise if length were taken as 4x (incorrect interpretation). Exam tip: always express length in terms of x first (x+4 here), multiply to form the quadratic, then bring 45 to one side before simplifying.

Factor the left side: \(x^2-8x=x(x-8)\). From \(x(x-8)=0\) we get \(x=0\) or \(x=8\), so the roots are 0 and 8. Option B is incorrect because solving \(x-8=0\) gives +8, not -8. Quick exam tip: if the constant term is 0, one root is 0; find the other by factoring or use sum/product of roots (sum = 8, product = 0).

The direct answer is option A: \\(x=\\pm\\frac{5}{3}\\). Begin with \\(9x^2-25=0\\). Add 25 to both sides: \\(9x^2=25\\). Divide by 9: \\(x^2=\\frac{25}{9}\\). Taking square roots gives \\(x=\\pm\\sqrt{\\frac{25}{9}}=\\pm\\frac{5}{3}\\). Both positive and negative values are required because both square to the same positive number. Option A is correct. Option B, \\(\\pm\\frac{3}{5}\\), reverses the numerator and denominator. Option C, \\(\\pm5\\), ignores the coefficient 9 of \\(x^2\\). Option D, \\(\\pm3\\), also ignores the constant 25 and does not satisfy the equation. Checking confirms that for \\(x=5/3\\), \\(9x^2-25=25-25=0\\), and the negative root also works. Memory cue: for \\(a^2x^2-b^2=0\\), roots are \\(x=\\pm b/a\\).

To factorise the quadratic we look for two numbers whose product is 12 (the constant term) and whose sum is 7 (the coefficient of x). The pair 3 and 4 satisfy this, so \((x+3)(x+4)=0\) is the correct factorisation. By the zero-product property the roots are x = −3 and x = −4. Closest distractor (B) yields a sum of −7 (wrong sign), (C) yields sum 8, and (D) gives product −12 — hence they are incorrect. Exam tip: match pairs whose product equals c and whose sum equals b to factor quickly for integer-coefficient quadratics.

From the standard form \(ax^2+bx+c=0\), we identify \(a=2,\ b=7,\ c=3\). Therefore \(ac=a\times c=2\times3=6\). Options B and D represent \(b\) and \(c\) respectively, so they are incorrect; option C (10) is not the product but an incorrect sum/product. Exam tip: first identify \(a, b, c\) from the equation, then compute the required combination (here multiply \(a\) and \(c\)).

For the general quadratic \(ax^2+bx+c=0\), \(b\) is the coefficient of \(x\) and \(c\) is the constant term. A missing \(x\)-term means its coefficient is \(0\). In \(3x^2-12=0\) we have \(b=0\) and \(c=-12\), and since \(-12<0\) this satisfies the requirement, so option B is correct. Closest distractor C (\(2x^2+4=0\)) also has \(b=0\) but \(c=4>0\), so it fails the \(c<0\) condition; option D has \(b=-1\) (not zero). Exam tip: rewrite each equation as \(ax^2+bx+c=0\), read off \(a,b,c\), and treat missing terms as zero to avoid mistakes.

The direct answer is option A, \(x^2+6x-10=0\). To remove the fraction, multiply every term and both sides of the equation by the denominator 2: \(2\times(x^2/2)+2\times3x-2\times5=2\times0\). This gives \(x^2+6x-10=0\). Multiplying an equation by a non-zero number does not change its solutions. Option A is correct. Option B, \(x^2+3x-5=0\), leaves the fractional coefficient effectively unchanged and does not result from multiplying every term by 2. Option C, \(2x^2+3x-5=0\), multiplies only the first term, not the whole equation, so it changes the equation incorrectly. Option D, \(x^2+6x+10=0\), has the wrong sign for the constant term; multiplying −5 by 2 gives −10, not +10. Exam cue: when clearing a denominator, multiply every term, including the constant.

To remove fractional coefficients multiply the entire equation by the LCM of denominators (here 3). Multiplying by 3 gives: \(3\times\frac{1}{3}x^2=x^2,\;3\times(-\frac{2}{3}x)=-2x,\;3\times1=3\). Thus the equivalent equation is \(x^2-2x+3=0\). Option C is wrong because it incorrectly applies the factor 3 (only the first term appears scaled); option B has the wrong sign for the x‑term and option D has the wrong constant. Exam tip: always multiply the whole equation by the LCM of denominators to obtain integer coefficients in one step.

In (x+\frac{1}{x}=2), the variable is in the denominator, so it is not directly in standard quadratic form. A quadratic polynomial form has no negative power.

The direct answer is option A: \\(b=2\\sqrt{3}\\). A quadratic equation in standard form is \\(ax^2+bx+c=0\\). Compare the given equation \\(x^2+2\\sqrt{3}x+3=0\\) with that pattern. The coefficient of \\(x^2\\) is 1, the coefficient of \\(x\\) is \\(2\\sqrt{3}\\), and the constant term is 3. Therefore \\(a=1\\), \\(b=2\\sqrt{3}\\), and \\(c=3\\). Option A is correct. Option B, \\(\\sqrt{3}\\), leaves out the multiplier 2. Option C, 3, is the constant term and is therefore \\(c\\), not \\(b\\). Option D, 1, is the coefficient of \\(x^2\\), so it is \\(a\\), not \\(b\\). The radical \\(\\sqrt{3}\\) is part of the coefficient and must not be ignored. Memory cue: in \\(ax^2+bx+c\\), read b as the complete coefficient attached to x.