Methods of Solving Quadratic Equations

(5x^2-13x-6=(5x+2)(x-3)), so the roots are (3) and (-\frac{2}{5}). In exams, reverse the signs from linear factors to write roots.

Bringing (6) to the left gives (3x^2+7x-6=0). In exams, make standard form before applying any method.

(3x^2+7x-6=(3x-2)(x+3)), so the roots are (\frac{2}{3}) and (-3). In exams, correct standard form is necessary first.

The governing identity is (p − q)² = p² − 2pq + q². Compare the expression with this pattern. The first term 16x² is (4x)², and the last term 9 is 3². The middle term produced by (4x − 3)² is −2(4x)(3) = −24x, exactly matching the given term. Therefore 16x² − 24x + 9 = (4x − 3)², so the equation becomes (4x − 3)² = 0 and option A is correct. Option B gives a positive middle term, while options C and D have incorrect leading terms and cannot expand to the original quadratic.

((4x-3)^2=0), so (4x-3=0) and (x=\frac{3}{4}). In exams, write the repeated root as a correct fraction.

(6x^2-18x=6x(x-3)), so (x=0) and (x=3) are both roots. In exams, dividing by the variable can miss (x=0).

Comparing the equation with (ax^2+bx+c=0 ), we get (a=3,b=-6,c=-2 ). Thus, (D=b^2-4ac=(-6)^2-4(3)(-2)=36+24=60 ). Since (c ) is negative, the term (-4ac ) becomes positive, so 24 must be added to 36 rather than subtracted. Exam tip: identify (a,b,c ) with their signs before applying the formula.

The formula gives (x=\frac{6\pm\sqrt{60}}{6}=1\pm\frac{\sqrt{15}}{3}). In exams, simplify (\sqrt{60}=2\sqrt{15}).

For equal roots, (k^2-100=0), so (k=\pm10), and (k<0) gives (k=-10). In exams, apply the given condition.

The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.

Using the identity \(a^2-2ab+b^2=(a-b)^2\), we get \(25x^2-20x+4=(5x)^2-2(5x)(2)+2^2=(5x-2)^2\). Hence, the correct perfect-square form is \((5x-2)^2=0\). In option B, the middle term would be \(+20x\), while options C and D have incorrect leading terms, \(625x^2\) and \(x^2\), respectively. Exam tip: Always check the sign and coefficient of the middle term when applying a perfect-square identity.

((5x-2)^2=0), so (5x-2=0) and (x=\frac{2}{5}). In exams, solve the linear equation after square form.

The standard form of a quadratic equation is ax² + bx + c = 0. Moving 7x to the left changes its sign to −7x, giving 2x² − 7x + 1 = 0. Therefore, option A is correct. Exam tip: the sign of a term changes whenever it is transposed to the other side of the equation.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=2, b=-7, c=1\\), so \\(D=(-7)^2-4(2)(1)=49-8=41\\). Option B results from taking only \\(b^2=49\\) and not subtracting \\(4ac\\). In an exam, identify \\(a,b,c\\) first and substitute their signed values carefully.

Adding (64) to (x^2-16x=-28) gives ((x-8)^2=36). In exams, add the square of half the coefficient.

Factoring the equation gives \(x^2-16x+28=(x-2)(x-14)\). Thus, \(x-2=0\) or \(x-14=0\), so the roots are \(x=2,14\). In option D, the sum of the numbers is 14, whereas the sum of the roots must be 16. As an exam tip, verify the roots using their sum \(16\) and product \(28\).

By Vieta’s sum-of-roots relation, the sum of the roots is \(-p\). Here, \((-4)+(-5)=-9\), so \(-p=-9\) and hence \(p=9\). Option B results from incorrectly taking the root sum as \(p\) instead of \(-p\). In exams, remember that for \(x^2+px+q=0\), the sum of the roots is \(-p\).

If roots are (3) and (7), then ((x-3)(x-7)=0), that is (x^2-10x+21=0). In exams, form factors with opposite signs of roots.

(3x^2+11x+10=(3x+5)(x+2)), so the roots are (-\frac{5}{3}) and (-2). In exams, positive factors give negative roots.

(x^2-x-30=(x-6)(x+5)), so the roots are (6) and (-5). In exams, the larger value decides the sign of the middle term.