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Class 11 Mathematics : Questions based on quadratic equations: Q4 – Q6

Q4 Find k so that the quadratic equation x2−(k+3)x+k=0 has equal and real roots.

Solution:

Step1 – For a quadratic ax2+bx+c=0 here a=1,  b=−(k+3),  c=k.

Step 2 — Equal (repeated) real roots occur when the discriminant Δ=b2−4ac equals zero. Compute:

Δ=(−(k+3))2−4⋅1⋅k=(k+3)2−4k.

Δ=k2+6k+9−4k=k2+2k+9.

Step 3 — Solve Δ=0. Set k2+2k+9=0. The discriminant of this quadratic in k is 22−4⋅1⋅9=4−36=−32<0, so there are no real solutions for k.

Conclusion: There is no real value of k that makes the original quadratic have equal real roots. (If complex k were allowed, k=−1±2i2.)


Q.5 Form a quadratic equation with integral coefficients whose roots are 2+5 and 2−5.

Solution:

Step 1 — If the roots are r1=2+5 and r2=2−5, then

sum =r1+r2=(2+5)+(2−5)=4,

product =r1r2=(2+5)(2−5)=4−5=−1.

Step 2 — A monic quadratic equation with these roots is

x2−(sum)x+(product)=x2−4x−1.Answer: quadratic equation is x2−4x−1=0. Its coefficients are integer.


Q.6 For roots α,β of 3x2−7x+2=0, compute α2+β2.

Solution:

Step 1 — For ax2+bx+c=0 we have sum= α+β=−ba and product = αβ=ca. Here a=3,  b=−7,  c=2, so

α+β=−−73=73,αβ=23.

Step 2 — By identity.

α2+β2=(α+β)2−2αβ.

Substitute the values:

α2+β2=(73)2−2⋅23=499−43.

convert to a common denominator:

499−129=379.

Answer : α2+β2=379.

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