Methods of Solving Quadratic Equations

(6x^2-7x-3=(3x+1)(2x-3)), so the roots are (-\frac{1}{3}) and (\frac{3}{2}). In exams, solve both linear factors separately.

Here (ac=72) and (-9+(-8)=-17), so the correct split is (-9x-8x). In exams, check both sum (b) and product (ac).

(12x^2-17x+6=(3x-2)(4x-3)), so the roots are (\frac{2}{3}) and (\frac{3}{4}). In exams, write fractional roots in simplest form.

For equal roots, (D=0), so (4(k+1)^2-4k^2=0) gives (2k+1=0). In exams, write (D) first in parameter questions.

The sum of roots is (2+4=6), and (\frac{p+3}{2}=6), so (p=9). In exams, use (-\frac{b}{a}) for sum of roots.

First we get \(x^2-\frac{10}{3}x+1=0\), then \(\left(x-\frac{5}{3}\right)^2=\frac{16}{9}\). In exams, divide by (a) first when \(a\neq1\).

(3x^2-10x+3=(3x-1)(x-3)), so the roots are (\frac{1}{3}) and (3). In exams, you may verify by completing square or factoring.

For real and equal roots, (D=0), so (36-4m=0) gives (m=9). In exams, remember equal roots mean (D=0).

The roots of the first equation are (2,\frac{1}{2}), and the roots of the second are (2,\frac{2}{3}). In exams, solve both equations separately for common root.

The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.

Since 2 is a root, substitute \(x=2\) in the equation: \(2^2-5(2)+q=0\), giving \(4-10+q=0\) and hence \(q=6\). Alternatively, the sum of the roots is 5, so the other root is 3 and their product is \(q=2\times3=6\). Exam tip: for \(x^2+bx+c=0\), the product of the roots is \(c\); 10 is related to neither the required constant term nor the correct substitution result.

(2x^2+3x-2=(2x-1)(x+2)), so (x=\frac{1}{2},-2) is correct. In exams, change signs carefully from factors.

Since (3=(\sqrt{3})^2) and the middle term is (2\sqrt{3}x), it is ((x+\sqrt{3})^2). In exams, identify perfect squares even with irrational coefficients.

The direct answer is option A: the root is \(x=-\sqrt{3}\), repeated twice. Start with \(x^2+2\sqrt{3}x+3\). Since \((\sqrt{3})^2=3\) and the middle term is twice \(x\) times \(\sqrt{3}\), the expression becomes \((x+\sqrt{3})^2=0\). Therefore \(x+\sqrt{3}=0\), so \(x=-\sqrt{3}\). Option A is correct because it gives this repeated root. Option B, \(x=\sqrt{3}\), has the wrong sign; substituting it gives a positive expression, not zero. Option C, \(x=-3\), confuses \(\sqrt{3}\) with 3 and is not a solution. Option D, \(x=3\), has both the wrong sign and wrong value. Memory cue: \((x+a)^2=0\) always gives the repeated root \(x=-a\).

The direct answer is option A: the roots are \(x=a+b\) and \(x=a-b\). Start with \(x^2-2ax+a^2-b^2=0\). The first three terms form \((x-a)^2\), so the equation becomes \((x-a)^2-b^2=0\). This is a difference of squares: \(((x-a)-b)((x-a)+b)=0\). Therefore either \(x-a=b\), giving \(x=a+b\), or \(x-a=-b\), giving \(x=a-b\). Option A is correct because it states both roots. Option B gives expressions beginning with \(-a\), which do not follow from the factorisation. Option C lists \(b+a\), which is the same as \(a+b\), but \(b-a\) is generally not the same as \(a-b\); it misses the required sign. Option D gives \(a^2\) and \(b^2\), but squaring the constants is not the method for finding these roots. Remember: when \(u^2-v^2=0\), write \(u=\pm v\).

Dividing the whole equation by (4) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first is easier.

Since (x^4=(x^2)^2=y^2), the new equation is (y^2-5y+4=0). In exams, substitution can simplify difficult forms.

From (y^2-5y+4=0), (y=1,4), so (x^2=1,4) and (x=\pm1,\pm2). In exams, do not forget to return to (x).

Multiplying both sides by (2x) gives (2+2x^2=5x), that is (2x^2-5x+2=0). In exams, remember the condition (x\neq0).

(2x^2-5x+2=(2x-1)(x-2)), so (x=\frac{1}{2}) and (2). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+1)^2=6x), so (x^2+2x+1=6x), and the correct form is (x^2-4x+1=0). In exams, cross multiply very carefully.

The direct answer is option A: \(x^2-4x+1=0\). The original fractions are defined only when \(x\neq0,-1\), so multiplying by \(x(x+1)\) is allowed under the stated restriction. Starting with \(\frac{x+1}{x}=\frac{6}{x+1}\), cross-multiply: \((x+1)(x+1)=6x\). Expand the square: \(x^2+2x+1=6x\). Move everything to the left: \(x^2+2x+1-6x=0\), hence \(x^2-4x+1=0\). Option A matches this exact result. Option B has the wrong middle coefficient: it would require \(+2x\) after simplification, which does not happen. Option C incorrectly keeps the number 6 as the middle coefficient and forgets the \(2x\) produced by expanding \((x+1)^2\). Option D has the wrong sign for the middle term. The excluded values remain excluded in the original equation, even if an algebraic transformation ever produces them. Memory cue: cross-multiply first, expand the square, then combine the middle terms.

(D=(-4)^2-4(1)(1)=12), so (x=\frac{4\pm2\sqrt{3}}{2}=2\pm\sqrt{3}). In exams, simplify the square root.

The sum of roots is (-\frac{p}{3}), so (-\frac{p}{3}=-5) gives (p=15). In exams, remember the sum formula (-\frac{b}{a}).

The product of roots is (\frac{p}{4}), so (\frac{p}{4}=\frac{3}{2}) gives (p=6). In exams, use the product formula (\frac{c}{a}).