Find the Values of k for Which kx² + 2x + 1 = 0 Has Real and Distinct Roots Find the Values of k for Which the Equation Has Real and Distinct Roots Solution Given: $$kx^2+2x+1=0$$ Here, $$a=k,\quad b=2,\quad c=1$$ For real and distinct roots, $$D=b^2-4ac>0$$ $$2^2-4(k)(1)>0$$ $$4-4k>0$$ $$1-k>0$$ $$k
Find the Values of k for Which 4x² + kx + 3 = 0 Has Equal Roots Find the Values of k for Which the Equation Has Equal Roots Solution Given: $$4x^2+kx+3=0$$ Here, $$a=4,\quad b=k,\quad c=3$$ For equal roots, $$D=b^2-4ac=0$$ $$k^2-4(4)(3)=0$$ $$k^2-48=0$$ $$k=\pm\sqrt{48}=\pm4\sqrt3$$ Answer The value(s) of k for which the roots are equal is:
Find the Values of k for Which kx(x − 3) + 9 = 0 Has Equal Roots Find the Values of k for Which the Equation Has Equal Roots Solution Given: $$kx(x-3)+9=0$$ $$kx^2-3kx+9=0$$ Here, $$a=k,\quad b=-3k,\quad c=9$$ For equal roots, $$D=b^2-4ac=0$$ $$(-3k)^2-4(k)(9)=0$$ $$9k^2-36k=0$$ $$9k(k-4)=0$$ $$k=0 \quad \text{or} \quad k=4$$ Since the equation must remain quadratic,
Find the Values of k for Which kx(x − 2√5) + 10 = 0 Has Equal Roots Find the Values of k for Which the Equation Has Equal Roots Solution Given: $$kx(x-2\sqrt5)+10=0$$ $$kx^2-2\sqrt5\,kx+10=0$$ Here, $$a=k,\quad b=-2\sqrt5\,k,\quad c=10$$ For equal roots, $$D=b^2-4ac=0$$ $$(-2\sqrt5\,k)^2-4(k)(10)=0$$ $$20k^2-40k=0$$ $$20k(k-2)=0$$ $$k=0 \quad \text{or} \quad k=2$$ Since the equation must remain quadratic,
Find the Values of k for Which x² − 4kx + k = 0 Has Equal Roots Find the Values of k for Which the Equation Has Equal Roots Solution Given: $$x^2-4kx+k=0$$ Here, $$a=1,\quad b=-4k,\quad c=k$$ For equal roots, $$D=b^2-4ac=0$$ $$(-4k)^2-4(1)(k)=0$$ $$16k^2-4k=0$$ $$4k(4k-1)=0$$ $$k=0 \quad \text{or} \quad k=\frac{1}{4}$$ Answer The value(s) of k for which
Find the Values of k for Which kx(x − 2) + 6 = 0 Has Equal Roots Find the Values of k for Which the Equation Has Equal Roots Solution Given: $$kx(x-2)+6=0$$ $$kx^2-2kx+6=0$$ Here, $$a=k,\quad b=-2k,\quad c=6$$ For equal roots, $$D=b^2-4ac=0$$ $$(-2k)^2-4(k)(6)=0$$ $$4k^2-24k=0$$ $$4k(k-6)=0$$ $$k=0 \quad \text{or} \quad k=6$$ Since the equation must remain quadratic,
Find the Values of k for Which 2x² + kx + 3 = 0 Has Equal Roots Find the Values of k for Which the Equation Has Equal Roots Solution Given: $$2x^2+kx+3=0$$ Here, $$a=2,\quad b=k,\quad c=3$$ For equal roots, $$D=b^2-4ac=0$$ $$k^2-4(2)(3)=0$$ $$k^2-24=0$$ $$k=\pm\sqrt{24}=\pm2\sqrt6$$ Answer The value(s) of k for which the roots are equal is:
Find the Values of k for Which x² + k(2x + k − 1) + 2 = 0 Has Real and Equal Roots Find the Values of k for Which the Equation Has Real and Equal Roots Solution Given: $$x^2+k(2x+k-1)+2=0$$ $$x^2+2kx+(k^2-k+2)=0$$ Here, $$a=1,\quad b=2k,\quad c=k^2-k+2$$ For real and equal roots, $$D=b^2-4ac=0$$ $$ (2k)^2-4(k^2-k+2)=0 $$ $$
Find the Values of k for Which (k + 1)x² − 2(k − 1)x + 1 = 0 Has Real and Equal Roots Find the Values of k for Which the Equation Has Real and Equal Roots Solution Given: $$ (k+1)x^2-2(k-1)x+1=0 $$ Here, $$ a=k+1,\quad b=-2(k-1),\quad c=1 $$ For real and equal roots, $$ D=b^2-4ac=0
Find the Values of k for Which k²x² − 2(2k − 1)x + 4 = 0 Has Real and Equal Roots Find the Values of k for Which the Equation Has Real and Equal Roots Solution Given: $$k^2x^2-2(2k-1)x+4=0$$ Here, $$a=k^2,\quad b=-2(2k-1),\quad c=4$$ For real and equal roots, $$D=b^2-4ac=0$$ $$[-2(2k-1)]^2-4(k^2)(4)=0$$ $$4(2k-1)^2-16k^2=0$$ $$ (2k-1)^2-4k^2=0 $$ $$ 4k^2-4k+1-4k^2=0