Here \(a=2\), \(b=-3\), \(c=4\).

The discriminant is

\(\Delta=b^2-4ac=(-3)^2-4(2)(4)=9-32=-23\).

Since \(\Delta<0\), the equation has no real roots.

\(a=5\), \(b=-11\), \(c=4\).

\(\Delta=(-11)^2-4(5)(4)=121-80=41\).

Since \(\Delta>0\), there are two distinct real roots.

\(a=1\), \(b=12\), \(c=36\).

\(\Delta=12^2-4(1)(36)=144-144=0\).

Since \(\Delta=0\), the equation has equal roots.

\(a=2\), \(b=1\), \(c=-2\).

\(\Delta=1^2-4(2)(-2)=1+16=17\).

Since \(\Delta>0\), there are two distinct real roots.

\(a=4\), \(b=5\), \(c=2\).

\(\Delta=5^2-4(4)(2)=25-32=-7\).

Since \(\Delta<0\), there are no real roots.

\(a=4\), \(b=-12\), \(c=9\).

\(\Delta=(-12)^2-4(4)(9)=144-144=0\).

So the roots are equal.

\(a=7\), \(b=-2\), \(c=1\).

\(\Delta=(-2)^2-4(7)(1)=4-28=-24\).

Since \(\Delta<0\), there are no real roots.

\(a=1\), \(b=-6\), \(c=7\).

\(\Delta=(-6)^2-4(1)(7)=36-28=8\).

Since \(\Delta>0\), there are two distinct real roots.

First write it in standard form:

\(-4x^2-x+2=0\).

Multiplying by \(-1\) gives \(4x^2+x-2=0\), which has the same roots.

\(\Delta=1^2-4(4)(-2)=1+32=33\).

Since \(\Delta>0\), there are two distinct real roots.

Write it as \(6x^2-6x+1=0\).

\(\Delta=(-6)^2-4(6)(1)=36-24=12\).

Since \(\Delta>0\), the equation has two distinct real roots.

Write it as \(16x^2-24x+9=0\).

\(\Delta=(-24)^2-4(16)(9)=576-576=0\).

So the roots are equal.

First expand:

\((x+2)^2-8x-2=x^2+4x+4-8x-2=x^2-4x+2\).

\(\Delta=(-4)^2-4(1)(2)=16-8=8\).

Since \(\Delta>0\), there are two distinct real roots.

\(\Delta=4^2-4(3)(5)=16-60=-44\).

Since \(\Delta<0\), there are no real roots.

\(\Delta=(-12)^2-4(9)(4)=144-144=0\).

So the roots are equal.

\(\Delta=2^2-4(1)(5)=4-20=-16\).

Since \(\Delta<0\), there are no real roots.

\(\Delta=13^2-4(6)(6)=169-144=25\).

Since \(\Delta>0\), there are two distinct real roots.

\(\Delta=(-4)^2-4(8)(1)=16-32=-16\).

Since \(\Delta<0\), there are no real roots.

\(\Delta=(-14)^2-4(1)(49)=196-196=0\).

So the roots are equal.

\(\Delta=5^2-4(12)(-2)=25+96=121\).

Since \(\Delta>0\), there are two distinct real roots.

Write it as \(x^2-10x+3=0\).

\(\Delta=(-10)^2-4(1)(3)=100-12=88\).

So there are two distinct real roots.

Write it as \(20x^2-20x+5=0\).

\(\Delta=(-20)^2-4(20)(5)=400-400=0\).

So the roots are equal.

\(\Delta=4^2-4(1)(8)=16-32=-16\).

Since \(\Delta<0\), there are no real roots.

Here \(A=2\), \(B=-a\), \(C=8\).

\(\Delta=(-a)^2-4(2)(8)=a^2-64\).

For no real roots, \(\Delta<0\).

So \(a^2-64<0\), hence \(a^2<64\).

Therefore \(-8<a<8\).

Write the equation in standard form:

\(kx^2+2x-6=0\).

For a repeated root, \(\Delta=0\).

\(\Delta=2^2-4(k)(-6)=4+24k\).

Set this equal to 0:

\(4+24k=0\Rightarrow k=-\dfrac16\).

For equal roots, \(\Delta=0\).

\(\Delta=p^2-4(1)(16)=p^2-64\).

So \(p^2-64=0\Rightarrow p^2=64\).

Hence \(p=\pm8\).

\(\Delta=6^2-4(1)(m)=36-4m\).

For no real roots, \(\Delta<0\).

So \(36-4m<0\Rightarrow 9-m<0\Rightarrow m>9\).

For a quadratic equation, we need \(k\neq0\).

The discriminant is

\(\Delta=5^2-4(k)(2)=25-8k\).

For two distinct real roots, \(\Delta>0\).

So \(25-8k>0\Rightarrow k<\dfrac{25}{8}\).

Also \(k\neq0\) because the equation must remain quadratic.

For equal roots, \(\Delta=0\).

Here \(A=p\), \(B=-4p\), \(C=2-p\).

\(\Delta=(-4p)^2-4(p)(2-p)=16p^2-8p+4p^2=20p^2-8p\).

So \(20p^2-8p=0\Rightarrow 4p(5p-2)=0\).

Since the equation is quadratic, \(p\neq0\).

Therefore \(5p-2=0\Rightarrow p=\dfrac25\).

\(\Delta=5^2-4(q-5)(-q)=25+4q(q-5)\).

Simplify:

\(\Delta=4q^2-20q+25=(2q-5)^2\).

Since a square is always non-negative, the equation has real roots for every real \(q\) (and when \(q=5\), the equation becomes linear with a real root).

Multiply through by \(x\):

\(x^2+kx+9=0\).

For equal roots, \(\Delta=0\).

\(\Delta=k^2-4(1)(9)=k^2-36\).

So \(k^2-36=0\Rightarrow k=\pm6\).

\(\Delta=4^2-4(a)(a)=16-4a^2\).

For no real roots, \(\Delta<0\).

So \(16-4a^2<0\Rightarrow 4-a^2<0\Rightarrow a^2>4\).

Hence \(a<-2\) or \(a>2\).

For equal roots, \(\Delta=0\).

\(\Delta=[2(k-1)]^2-4(1)(9)=4(k-1)^2-36\).

So \(4(k-1)^2=36\Rightarrow (k-1)^2=9\).

Hence \(k-1=\pm3\), so \(k=4\) or \(k=-2\).

\(\Delta=m^2-4(3)(12)=m^2-144\).

For two distinct real roots, \(\Delta>0\).

So \(m^2-144>0\Rightarrow m^2>144\).

Hence \(m<-12\) or \(m>12\).

\(\Delta=t^2-4(2)(18)=t^2-144\).

For no real roots, \(\Delta<0\).

So \(t^2<144\).

Hence \(-12<t<12\).

\(\Delta=(k+1)^2-4(1)(k)=k^2-2k+1=(k-1)^2\).

For equal roots, \(\Delta=0\).

So \((k-1)^2=0\), which gives \(k=1\).

\(\Delta=6^2-4(p-1)p=36-4p^2+4p\).

For no real roots, \(\Delta<0\).

So \(36-4p^2+4p<0\).

Divide by 4:

\(9-p^2+p<0\Rightarrow p^2-p-9>0\).

The roots of \(p^2-p-9=0\) are \(p=\dfrac{1\pm\sqrt{37}}{2}\).

Therefore \(p<\dfrac{1-\sqrt{37}}{2}\) or \(p>\dfrac{1+\sqrt{37}}{2}\).

\(\Delta=(-4k)^2-4(4)(k^2)=16k^2-16k^2=0\).

So the discriminant is zero for every real value of \(k\).

Therefore the equation has equal roots for all real \(k\).

Because the equation is quadratic, \(m\neq0\).

\(\Delta=(-6)^2-4(m)(m)=36-4m^2\).

For two distinct real roots, \(\Delta>0\).

So \(36-4m^2>0\Rightarrow 9-m^2>0\Rightarrow -3<m<3\).

Combining with \(m\neq0\), we get \(-3<m<3,\ m\neq0\).

\(\Delta=8^2-4(a)(2)=64-8a\).

For no real roots, \(\Delta<0\).

So \(64-8a<0\Rightarrow a>8\).

For equal roots, \(\Delta=0\).

\(\Delta=b^2-4(3)(3)=b^2-36\).

So \(b^2-36=0\Rightarrow b=\pm6\).

For \(x^2+px+q=0\), the sum of the roots is \(-p\) and the product is \(q\).

Here the sum is \(-3+2=-1\), so \(-p=-1\), giving \(p=1\).

The product is \((-3)(2)=-6\), so \(q=-6\).

The monic quadratic is

\(x^2-\left(-\dfrac12+\dfrac34\right)x+\left(-\dfrac12\cdot\dfrac34\right)=0\).

The sum of the roots is \(\dfrac14\), and the product is \(-\dfrac38\).

So the equation is

\(x^2-\dfrac14x-\dfrac38=0\).

Multiply through by 8 to get integer coefficients:

\(8x^2-2x-3=0\).

Therefore \(p=8\), \(q=-2\), \(r=-3\).

Multiply through by \(x\):

\(3x^2+4x-5=0\).

The discriminant is

\(\Delta=4^2-4(3)(-5)=16+60=76\).

Since \(\Delta>0\), there are two distinct real roots.

Complete the square:

\(x^2+8x+16=(x+4)^2\).

A square is always non-negative, so \(x^2+8x+16\ge0\) for all real \(x\).

The minimum value is 0, which occurs when \(x=-4\).

This also means the equation \(x^2+8x+16=0\) has equal roots.

Rearrange the terms:

\(100x^2-20x+1=(10x-1)^2\).

So \(1+100x^2-20x\ge0\) for all real \(x\).

Its minimum value is 0 when \(x=\dfrac{1}{10}\).

Therefore the equation \(1+100x^2-20x=0\) has equal roots.

The sum of the roots is \(4+(-7)=-3\), so \(-p=-3\), which gives \(p=3\).

The product is \(4(-7)=-28\), so \(q=-28\).

For equal roots, \(\Delta=0\).

\(\Delta=p^2-4(2)(8)=p^2-64\).

So \(p^2=64\), hence \(p=\pm8\).

\(\Delta=p^2-4(7)(2)=p^2-56\).

For no real roots, \(\Delta<0\).

So \(p^2<56\).

Hence \(-2\sqrt{14}<p<2\sqrt{14}\).

For equal roots, \(\Delta=0\).

\(\Delta=4^2-4(k)(1)=16-4k\).

So \(16-4k=0\Rightarrow k=4\).

Multiply through by \(x\):

\(x^2+x-4=0\).

The discriminant is

\(\Delta=1^2-4(1)(-4)=1+16=17\).

Since \(\Delta>0\), there are two distinct real roots.