Nature of Roots

Here, \(a=4\), \(b=4\), and \(c=5\). The discriminant is \(D=b^2-4ac=4^2-4(4)(5)=16-80=-64<0\). Hence, the equation has no real roots; its roots are complex. Exam tip: For a quadratic equation, \(D<0\) indicates no real roots.

For the given equation, \(a=1\), \(b=5\), and \(c=6\). Thus, \(\Delta=b^2-4ac=5^2-4(1)(6)=25-24=1\). Since \(\Delta>0\), the equation has two real and distinct roots. Option B would be correct only if \(\Delta=0\). In exams, calculate the discriminant first and then use its sign to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-2\), and \(c=4\), so \(D=(-2)^2-4(3)(4)=4-48=-44\). Since \(D<0\), the discriminant is negative and the equation has no real roots. It is not zero; a zero discriminant occurs only when \(b^2=4ac\). In an exam, first identify \(a,b,c\) and then apply \(D=b^2-4ac\).

Here, \(a=2\), \(b=4\), and \(c=2\). Thus, \(D=b^2-4ac=4^2-4(2)(2)=16-16=0\). Therefore, the discriminant is zero, and the equation has two equal real roots. Choosing positive or negative is incorrect. Exam tip: when \(D=0\), the roots are real and equal.

Here, \(a=1\), \(b=-11\), and \(c=30\). The discriminant is \(D=b^2-4ac=(-11)^2-4(1)(30)=121-120=1\). Since \(D>0\), the roots are real and distinct. In fact, the equation factors as \((x-5)(x-6)=0\), giving roots 5 and 6. Exam tip: \(D>0\) indicates real and distinct roots, \(D=0\) indicates real and equal roots, and \(D<0\) indicates non-real roots.

Here, \(a=1\), \(b=14\), and \(c=49\). Thus, the discriminant is \(D=b^2-4ac=14^2-4(1)(49)=196-196=0\). When \(D=0\), the two roots are real and equal. In fact, the equation can be written as \((x+7)^2=0\), giving the repeated root \(-7\). Exam tip: choose equal real roots for \(D=0\); distinct real roots require \(D>0\).

Here, \(a=1\), \(b=-4\), and \(c=8\). The discriminant is \(D=b^2-4ac=(-4)^2-4(1)(8)=16-32=-16\). Since \(D<0\), the equation has no real roots. In fact, its roots are \(2+2i\) and \(2-2i\), which are non-real and distinct. Exam tip: For a quadratic equation, \(D<0\) means that there are no real roots.

The governing condition for equal real roots of ax² + bx + c = 0 is a zero discriminant, b² − 4ac = 0. In x² − 3x + s = 0, the coefficients are a = 1, b = −3 and c = s. Hence (−3)² − 4(1)(s) = 0, which gives 9 − 4s = 0 and therefore s = 9/4. The repeated root would be −b/(2a) = 3/2, and substituting it confirms the result: (3/2)² − 3(3/2) + 9/4 = 0. Thus A is correct. Values such as 3/4 or 3 produce a nonzero discriminant and do not give equal roots.

For a quadratic equation \(ax^2+bx+c=0\) to have real and equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=5\), \(b=-4\), and \(c=t\), so \((-4)^2-4(5)t=16-20t=0\). Therefore, \(t=\frac{4}{5}\). Option B results from reversing the numerator and denominator. Exam tip: whenever roots are real and equal, immediately use \(D=0\).

For \(x^2+1=0\), \(a=1, b=0, c=1\). Thus, the discriminant is \(D=b^2-4ac=0^2-4(1)(1)=-4<0\). A quadratic equation with a negative discriminant has no real roots, so the statement is correct. Exam tip: For \(D<0\), write ‘no real roots.’

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=3, b=10, c=3\\), so \\(D=10^2-4(3)(3)=64\\). Since \\(D>0\\), the two roots are real and distinct. If \\(D=0\\), the roots would be real and equal, so option B is incorrect. Exam tip: The sign of the discriminant directly determines the nature of the roots.

The nature of the roots of a quadratic equation is determined by its discriminant, \(D=b^2-4ac\). If \(D>0\), the roots are real and distinct; if \(D=0\), the roots are real and equal; and if \(D<0\), there are no real roots. Therefore, the discriminant is the correct quantity. Exam tip: first write the equation in standard form and identify \(a\), \(b\), and \(c\) before calculating \(D\).

For a quadratic equation, the discriminant is \(D=b^2-4ac\). When \(D>0\), \(\sqrt{D}\) is positive and real, so the two signs in \(x=\frac{-b\pm\sqrt{D}}{2a}\) give two distinct real roots. Equal roots occur when \(D=0\), while no real roots occur when \(D<0\); therefore, options B and C are incorrect. In an exam, first check the sign of the discriminant to determine the nature of the roots.

The discriminant is \(D=b^2-4ac\). When \(D<0\), the root formula contains \(\sqrt{D}\), the square root of a negative number. Hence, there are no real roots; the roots form a complex conjugate pair. Exam tip: \(D>0\) gives two distinct real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-5\), and \(c=6\), so \(D=(-5)^2-4(1)(6)=25-24=1\). The value 0 would indicate equal roots, but this equation has \(D=1\). Exam tip: identify the signs of \(a\), \(b\), and \(c\) carefully before substituting.

Here, \(a=1, b=-4, c=4\). The discriminant is \(D=b^2-4ac=(-4)^2-4(1)(4)=0\), so the roots are real and equal. In fact, \(x^2-4x+4=(x-2)^2\), giving the repeated root \(x=2\). Exam tip: when \(D=0\), the roots are always real and equal.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=2\), \(b=-3\), and \(c=1\), so \(D=(-3)^2-4(2)(1)=9-8=1\). Therefore, the correct answer is 1. In exams, be careful to square the negative value of \(b\) correctly.

For a quadratic equation ax²+bx+c=0, the discriminant is D=b²−4ac. Here a=2, b=−3, and c=1, so D=(−3)²−4(2)(1)=9−8=1. Since D is positive, the equation has two real and distinct roots. In fact, factoring gives 2x²−3x+1=(2x−1)(x−1), so the roots are 1/2 and 1, confirming that they are unequal and real. Therefore option A is correct; D=0 would give equal roots, while D<0 would give non-real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3, b=6, c=3\), so \(D=6^2-4(3)(3)=0\). When \(D=0\), the roots are real and equal; in fact, both roots are \(-1\). Therefore, option A is correct. Exam tip: \(D=0\) always indicates equal real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1,b=1,c=1\), so \(D=1^2-4(1)(1)=-3\), which is negative. Therefore, option A is correct. The discriminant would be zero only if \(D=0\), which is not the case here. Exam tip: when \(D<0\), the quadratic equation has no real roots.