Nature of Roots

For the quadratic equation, \(a=1\), \(b=-6\), and \(c=10\). Thus, the discriminant is \(D=b^2-4ac=(-6)^2-4(1)(10)=36-40=-4\). Since \(D<0\), the equation has no real roots, so \(s\) cannot have a real value in the area problem. Exam tip: \(D<0\), \(D=0\), and \(D>0\) indicate no real roots, equal real roots, and distinct real roots, respectively.

For the given equation, \(a=1\), \(b=-15\), and \(c=56\). Therefore, the discriminant is \(D=b^2-4ac=(-15)^2-4(1)(56)=225-224=1\). Since \(D>0\) and 1 is a perfect square, the roots are real, rational, and distinct. Exam tip: a positive perfect-square discriminant gives two distinct rational roots; merely having \(D>0\) does not guarantee rational roots.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=1,b=8,c=18\\), so \\(D=8^2-4(1)(18)=64-72=-8\\). Since \\(D<0\\), the equation has no real roots. Option B would be correct only if \\(D=0\\). Exam tip: Always check the sign of the discriminant before identifying the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(D=(-12)^2-4(4)(11)=144-176=-32\). Since \(D<0\), the equation has no real roots. Option A is incorrect because a negative discriminant does not produce real irrational roots; it produces non-real complex roots. Exam tip: remember that \(D<0\) means no real roots.

Here, \(a=1, b=-13, c=40\). Therefore, the discriminant is \(D=b^2-4ac=(-13)^2-4(1)(40)=169-160=9\). Since \(D>0\), the roots are real and distinct; because \(D=9\) is a perfect square, they are also rational. In fact, the roots are \(5\) and \(8\). Exam tip: If \(D>0\) and is a perfect square, the roots are real, distinct, and rational.

Here (D=(-4)^2-4(3)(-1)=28). (28) is positive but not a perfect square, so the roots are irrational and distinct.

Here, \(a=2, b=-11, c=15\), so the discriminant is \(\Delta=b^2-4ac=(-11)^2-4(2)(15)=1\). Since \(\Delta\) is positive and a perfect square, the roots are real, rational, and distinct. In fact, \(2x^2-11x+15=(2x-5)(x-3)\), giving the roots \(x=\frac{5}{2}\) and \(x=3\). Exam tip: when \(\Delta>0\) and is a perfect square, the roots are rational and distinct.

Here, \(a=1, b=10, c=29\). Therefore, the discriminant is \(D=b^2-4ac=10^2-4(1)(29)=100-116=-16\). Since \(D<0\), the equation has no real roots. Option B is incorrect because equal real roots require \(D=0\). Exam tip: To determine the nature of roots, first calculate the discriminant and check its sign.

A quadratic equation \(ax^2+bx+c=0\) has real roots when its discriminant \(D=b^2-4ac\) is non-negative. Here, \(a=1, b=-2k, c=9\), so \(D=(-2k)^2-4(1)(9)=4k^2-36\). Thus, \(4k^2-36\geq0\), which gives \(k^2\geq9\), or \(k\leq-3\) or \(k\geq3\). Option B makes the discriminant negative, while option D is incomplete because \(k=-3\) also gives real equal roots. Exam tip: For real roots, always apply the condition \(D\geq0\).

For a quadratic equation ax²+bx+c=0, the discriminant is D=b²−4ac. Its sign determines the nature of the roots. When D>0, the square root of D is real and nonzero, so the quadratic formula gives two real and distinct roots: x=(-b+√D)/(2a) and x=(-b−√D)/(2a). If D is also a perfect square, these roots are rational, assuming rational coefficients in the usual school setting. However, because D is not a perfect square, √D is irrational, making both roots irrational. Hence option A is correct. Options B, C, and D correspond respectively to a square positive discriminant, zero discriminant, and negative discriminant.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1,b=-1,c=-1\), so \(D=(-1)^2-4(1)(-1)=5\). Since \(D>0\), the roots are real and distinct; since 5 is not a perfect square, the roots are irrational. Therefore, option A is correct. Exam tip: If \(D>0\) and \(D\) is not a perfect square, the roots are real, distinct, and irrational.

For the given equation, \(a=2\), \(b=-1\), and \(c=2\). Thus, the discriminant is \(D=b^2-4ac=(-1)^2-4(2)(2)=1-16=-15\). Since \(D<0\), the quadratic equation has no real roots. Exam tip: calculate the discriminant first and identify the nature of the roots from its sign.

The equation can be factorised as \(9x^2-6x+1=(3x-1)^2\). Thus, \((3x-1)^2=0\), giving \(3x-1=0\) and the equal root \(x=\frac{1}{3}\). Exam tip: For equal roots, the discriminant \(D=b^2-4ac\) is zero.

Here, \(a=4, b=7, c=-2\). Therefore, the discriminant is \(D=b^2-4ac=7^2-4(4)(-2)=81\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct. In fact, the roots are \(\frac{1}{4}\) and \(-2\). Option B is incorrect because a positive discriminant gives irrational roots only when it is not a perfect square. Exam tip: If \(D>0\) and is a perfect square, the roots are real, rational, and distinct.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=11, c=31\), so \(D=11^2-4(1)(31)=121-124=-3\). Since \(D<0\), the equation has no real roots. Exam tip: two real and equal roots occur only when \(D=0\), not when the discriminant is negative.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-2\), and \(c=-2\), so \(D=(-2)^2-4(3)(-2)=28\). Since \(D>0\), the roots are real and distinct; since 28 is not a perfect square, the roots are irrational. Option B is incorrect because the discriminant is 28, not 4. Exam tip: If \(D>0\) and is not a perfect square, the roots are real, distinct, and irrational.

Here, \\(a=16, b=24, c=9\\). The discriminant is \\(D=b^2-4ac=24^2-4(16)(9)=0\\), so the roots are real and equal. Also, \\(16x^2+24x+9=(4x+3)^2\\), which gives \\(4x+3=0\\) and hence the equal root \\(x=-\frac{3}{4}\\). Therefore, option B has the wrong sign, while option C gives an incorrect discriminant. Exam tip: If a quadratic is a perfect square or its discriminant is zero, its roots are equal.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=5, b=1, c=1\), so \(D=1^2-4(5)(1)=-19\). Since \(D<0\), the equation has no real roots; therefore, option A is correct. Exam tip: \(D<0\) means no real roots, \(D=0\) means equal real roots, and \(D>0\) means two distinct real roots.

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=6, b=-13, c=6\\), so \\(D=(-13)^2-4(6)(6)=169-144=25\\). Since \\(D>0\\) and 25 is a perfect square, the roots are real, rational, and distinct. In fact, the roots are \\(2/3\\) and \\(3/2\\). Option B would require \\(D=0\\). Exam tip: a positive perfect-square discriminant indicates distinct rational roots.

A quadratic equation (ax^2+bx+c=0) has equal roots when its discriminant (D=b^2-4ac) is zero. Here, (a=1,b=-14,c=k), so (-14)^2-4(1)(k)=0, giving (196-4k=0) and hence (k=49). The value 196 results from forgetting to divide by 4. Exam tip: For equal roots, immediately use (D=0).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here \(a=1\), \(b=k\), and \(c=36\), so \(D=k^2-4(1)(36)=k^2-144\). Therefore, \(k^2-144=0\), giving \(k^2=144\) and \(k=\pm12\). Exam tip: For equal roots of a quadratic equation, set the discriminant directly equal to zero.

Here, \(a=1\), \(b=-16\), and \(c=k\). Thus, the discriminant is \(D=b^2-4ac=(-16)^2-4(1)(k)=256-4k\). Two real and distinct roots require \(D>0\), so \(256-4k>0\), which gives \(k<64\). When \(k=64\), the roots are equal, so that option is not correct. Exam tip: Check the sign of the discriminant first when determining the nature of roots.

For this quadratic equation, \(a=1\), \(b=6\), and \(c=k\). Its discriminant is \(D=b^2-4ac=36-4k\). No real roots occur when \(D<0\), so \(36-4k<0\), which gives \(k>9\). At \(k=9\), the equation has one repeated real root, while \(k<9\) gives two real roots. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq0\). Here, \(a=3\), \(b=-4\), and \(c=p\), so \(D=b^2-4ac=(-4)^2-4(3)(p)=16-12p\). Thus, \(16-12p\geq0\), which gives \(p\leq\frac{4}{3}\). Hence, option A is correct. Option B reverses the required inequality, while \(p<0\) is only a sufficient condition, not the complete condition. Exam tip: For questions about real roots, begin by applying \(D\geq0\).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=2\), \(b=m\), and \(c=18\), so \(m^2-4(2)(18)=0\), giving \(m^2=144\). Therefore, \(m=\pm12\), or \(m=12\) or \(m=-12\). Option B results from an incorrect calculation of \(4ac\). Exam tip: whenever equal roots are mentioned, immediately use \(D=0\).