Methods of Solving Quadratic Equations

The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.

In (x^2-8x=0), (x) is a common factor and (x=0) can be a root. In exams, write (x(x-8)=0) in such cases.

Use the identity \(a^2-2ab+b^2=(a-b)^2\). Here, \(9x^2=(3x)^2\), \(25=5^2\), and the middle term is \(-30x=-2\cdot3x\cdot5\). Therefore, \(9x^2-30x+25=(3x-5)^2\), so the equation becomes \((3x-5)^2=0\). Option B would produce a positive middle term. In exams, take the square roots of the first and last terms and verify the middle term using \(\pm2ab\).

((3x-5)^2=0), so (3x-5=0) and (x=\frac{5}{3}). In exams, solve the linear equation after square form.

Bringing (8x) to the left gives (4x^2-8x-1=0). In exams, do not miss any term while making standard form.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=4\), \(b=-8\), and \(c=-1\), so \(D=(-8)^2-4(4)(-1)=64+16=80\). The value 48 results from incorrectly treating the contribution of \(-4ac\) as negative. Exam tip: when \(c\) is negative, \(-4ac\) contributes a positive value.

From (x^2-12x+20=0), (x^2-12x=-20), then adding (36) gives ((x-6)^2=16). In exams, add the same number to both sides.

Write the equation by completing the square: \(x^2-12x+20=(x-6)^2-16=0\). Thus, \((x-6)^2=16\), so \(x-6=\pm4\), giving \(x=2\) or \(x=10\). The values in option C have sum 12 but product 32, whereas the product of the roots must be 20. Exam tip: always include both values obtained from \(\pm\) in a quadratic equation.

The factors corresponding to the roots are \((x+3)(x+4)\). Expanding gives \(x^2+7x+12=0\), so \(p=7\). Option B incorrectly uses the sum of the roots, \(-7\), whereas the coefficient is its negative. Exam tip: for \(x^2+px+q=0\), the sum of the roots is \(-p\).

If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is (x^2-7x+10=0). In exams, form factors with opposite signs of roots.

(2x^2+7x+6=(2x+3)(x+2)), so (x=-\frac{3}{2}) and (-2). In exams, positive factors give negative roots.

(x^2-3x-10=(x-5)(x+2)), so the roots are (5) and (-2). In exams, in mixed signs the larger value decides the middle-term sign.

(D=(-6)^2-4(1)(8)=4), so (x=\frac{6\pm2}{2}) gives (2) and (4). In exams, keep the sign of (-b) correct.

(-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.

From (7x^2-28x=0), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.

To complete the square, first move the constant term to the other side: x² + 4x = 12. The coefficient of x is 4, so half of it is 2; adding its square, 2² = 4, creates a perfect square. Add 4 to both sides: x² + 4x + 4 = 12 + 4 = 16. The left side factors as (x + 2)², giving (x + 2)² = 16. Thus option A is correct. Option B uses the wrong sign, option C uses 4 rather than half of 4 inside the bracket and fails to preserve the equation, while option D adds 4 but forgets to add it to the right side.

Factor the equation as \\(x^2+6x-2x-12=0\\), giving \\( (x+6)(x-2)=0 \\). Therefore, \\(x=-6\\) or \\(x=2\\), so option A is correct. In option B, the signs of both roots are incorrect. As an exam check, the roots should have sum \\(-4\\) and product \\(-12\\).

(ac=-12) and (4+(-3)=1), so (x) is split as (4x-3x). In exams, check the sign of (ac) carefully.

The equal root is (x=\frac{-b}{2a}), so (x=\frac{8}{4}=2). In exams, this short formula is useful when (D=0).

Here (D=(-2)^2-4(1)(5)=-16<0), so there are no real roots. In exams, (D<0) means no real solution.