Introduction to Quadratic Equations

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).

(9x^2=25), so (x^2=\frac{25}{9}) and (x=\pm \frac{5}{3}). Take both signs while taking square roots.

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.

Multiplying the whole equation by (2) gives (x^2+6x-10=0). To remove fractions, multiply by the denominator.

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 coefficient attached to (x) is (2\sqrt{3}). Radical coefficients are treated like ordinary coefficients.

The sum of roots is (2+(-3)=-1). When adding integers with unlike signs, take the sign of the larger magnitude.

The product of roots is (4\cdot6=24). Sum and product are both useful while forming an equation from roots.

For a monic quadratic with root-sum S and product P the equation is \(x^2-Sx+P=0\). With S=5 and P=6 this gives \(x^2-5x+6=0\), so option A is correct. Check distractors: B has product 6 but sum \(-5\), while C and D have product 5, not 6. Exam tip: remember for \(ax^2+bx+c=0\) the sum of roots = \(-b/a)\) and product = \(c/a)\).

The sum of roots is (-\frac{b}{a}=-\frac{-5}{2}=\frac{5}{2}). Pay attention to the sign of (b) in the formula.

For a quadratic \(ax^2+bx+c=0\), the product of roots equals \(c/a\) (Vieta's formula). Here \(a=3\) and \(c=-8\), so the product is \(\dfrac{-8}{3}=-\tfrac{8}{3}\). Option B (\(\tfrac{8}{3}\)) is a sign error; options C and D have incorrect numerical values. Exam tip: apply \(c/a\) directly and pay attention to the sign of \(c\) from the given equation.

For a quadratic \(ax^2+bx+c=0\), the product of the roots equals \(\dfrac{c}{a}\). Here \(a=k\) and \(c=2\), so \(\dfrac{2}{k}=1\) which gives \(k=2\). Option 1 might arise from mistakenly taking \(c\) alone instead of \(c/a\); options -2 and 4 do not satisfy the relation \(2/k=1\). Exam tip: always identify \(a\) (coefficient of \(x^2\)) before applying the product formula \(c/a\).

Putting (x=1) gives (1+1+1=3\neq 0). To check when a value is not a root, use substitution too.

The sum of the roots is 4+4 = 8. For a quadratic \(ax^2+bx+c=0\), the sum of roots equals \(-\frac{b}{a}\). Here \(a=1, b=p\), so \(-p=8\) and hence \(p=-8\). The product check also matches: \(4\times4=16=c/a\). The common wrong choice 8 comes from forgetting the negative sign in the relation. Exam tip: use sum = -b/a and product = c/a and verify both to avoid sign errors.

To simplify the equation, first expand the square of the binomial. The identity \\(x+2)^2 = x^2 + 4x + 4\\) is used because the middle term is twice the product of x and 2. The remaining terms are then combined like ordinary algebraic terms. Care is needed because the expression contains subtraction of both (x+2) and 6.

Substituting the expansion gives x^2 + 4x + 4 - x - 2 - 6 = 0. Combining like terms produces x^2 + 3x - 4 = 0. Therefore option A is correct. The coefficient of x is 3 because 4x - x = 3x, and the constant is -4 because 4 - 2 - 6 = -4. The other options result from expanding or combining the terms incorrectly.

The left side gives (x^2+5x-6), and subtracting (2x) gives (x^2+3x-6=0). First expand and then bring all terms to one side.