Quadratic Equations Class 10 Exercise 4.3 Solutions

Let the discriminant do the deciding before any formula. The sign of D=b²-4ac settles the nature of the roots in a single line, so always compute it first, state the verdict, and reach for the roots only when it is non-negative.

• Square the surd fully: in part (ii) the middle coefficient carries a root of three, and squaring it gives a positive forty-eight, not a negative; forgetting that the surd also squares is the usual error that turns a zero discriminant into a wrong sign.
• Read off equal roots: when the discriminant is zero the two roots coincide at minus the middle coefficient over twice the leading one, and that single value may be left as a surd fraction or rationalised, since both forms are accepted.
• Simplify before dividing: in part (iii), reducing the root of twelve to two roots of three before dividing lets the common factor cancel cleanly and gives the neat half-form rather than an ugly unreduced answer.

No real roots; equal roots 2/√3; distinct roots (3±√3)/2.

Equal roots is the same as discriminant zero, but watch the leading coefficient. The phrase "two equal roots" should immediately trigger the condition that the discriminant equals zero, and the only skill is reading off the coefficients correctly when they carry the unknown.

• Both signs stand: in part (i) the condition gives a square equal to twenty-four, so the unknown is plus or minus two roots of six, and both signs are valid because either one makes the discriminant vanish exactly.
• Guard the leading term: part (ii) factors to offer a zero and a six, but the zero removes the squared term and leaves a false numerical statement, so it cannot be accepted, and that rejection, stated with a reason, is the single most marks-sensitive line here.
• Reduce the surd: simplify surd answers fully rather than leaving an unreduced root, since examiners expect the lowest form, and a half-simplified surd often costs the presentation mark.

k=2√6 for (i); k=6 for (ii).

Answer the yes-or-no part with the discriminant, then give the sizes. The question really has two halves, whether such a grove can exist at all and, if it can, what its measurements are, and each half deserves its own line.

• Settle existence first: the discriminant of the area equation comes out positive, so real roots exist and the honest answer to "is it possible" is yes, written as a clear sentence before any number appears.
• Then size it: solving the simplified equation gives a positive breadth, while the negative root is rejected because a length cannot be negative, and doubling the breadth gives the length.
• Why the discriminant line matters: the "is it possible" phrasing is the examiner's cue that the nature-of-roots idea is being tested, so writing that line explicitly earns the marks even though the algebra is simple enough to solve directly.

Possible: breadth 20 m, length 40 m (area 4020=800 m²).

A negative discriminant is the proof of impossibility. The two conditions translate into a single quadratic, and the temptation is to keep hunting for roots, but the discriminant turns out negative, so no real value exists and the situation simply cannot happen.

• Present it as a proof: compute the discriminant, note that it is negative, and write one sentence concluding the ages are impossible, because the marks here reward recognising the no-real-roots condition, not any numerical age.
• Build the second age right: the ages four years ago are x-4 and 16-x, so the second must use the sum of twenty first and only then subtract four, which is the step students most often slip on.
• Tidy the leading sign: after expanding, multiply through by minus one so the leading coefficient is positive, which lets the discriminant be read off cleanly without a sign mistake.

Not possible, since D=-48<0 gives no real ages.

Use perimeter to reduce to one unknown, then test with the discriminant. The two facts hand you a sum from the perimeter and a product from the area, so the length and breadth are simply the roots of the matching quadratic.

• Zero means equal: the discriminant comes out exactly zero, which signals both that a solution exists and that the two roots coincide, so the answer is yes and there is a single repeated value.
• Name the square: because length and breadth come out equal, the park is actually a square, and a student should say so plainly rather than treating the matching values as a mistake to be fixed.
• Verify both facts: the cleanest finish checks the perimeter and the area against the original numbers, and spotting the perfect-square form of the quadratic is an even faster route that also confirms the zero discriminant.

Possible: length = breadth =20 m, a square park.