Skip to content
Question 114 of 128

Q.If one root of k(x−1)2=5x−7k(x-1)^2 = 5x-7 is double the other root, show that k=2k=2 or −25-25. OR Express the matrix A=[135−683−465]A = \begin{bmatrix}1 & 3 & 5\\-6 & 8 & 3\\-4 & 6 & 5\end{bmatrix} as the sum of a symmetric and a skew symmetric matrices.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 5mImportance★★★★★
89% · 114/128 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Writing the roots as rr and 2r2r, using sum =3r=3r and product =2r2=2r^2 from the quadratic's coefficients, and eliminating rr leads to k2+23k−50=0k^2+23k-50=0, whose roots are k=2k=2 and k=−25k=-25.

Expand k(x−1)2=5x−7k(x-1)^2=5x-7:

k(x2−2x+1)=5x−7  ⟹  kx2−2kx+k−5x+7=0  ⟹  kx2−(2k+5)x+(k+7)=0k(x^2-2x+1)=5x-7 \implies kx^2-2kx+k-5x+7=0 \implies kx^2-(2k+5)x+(k+7)=0

Let the roots be rr and 2r2r (one root is double the other). By Vieta's formulas:

Sum: r+2r=3r=2k+5k  ⟹  r=2k+53k\text{Sum: } r+2r=3r=\frac{2k+5}{k} \implies r=\frac{2k+5}{3k}

Product: r(2r)=2r2=k+7k  ⟹  r2=k+72k\text{Product: } r(2r)=2r^2=\frac{k+7}{k} \implies r^2=\frac{k+7}{2k}

Square the sum expression and set equal to the product expression:

(2k+53k)2=k+72k\left(\frac{2k+5}{3k}\right)^2 = \frac{k+7}{2k}

(2k+5)29k2=k+72k\frac{(2k+5)^2}{9k^2} = \frac{k+7}{2k}

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.