Question

# Find the values of k for which the following equations have real roots (i) (ii) (iii) ${x}^{2}-4kx+k=0$ (iv) $kx\left(x-2\sqrt{5}\right)+10=0$ (v) $kx\left(x-3\right)+9=0$ (vi) $4{x}^{2}+kx+3=0$

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Solution

## (i) The given quadric equation is , and roots are real and equal Then find the value of k. Here, As we know that Putting the value of The given equation will have real and equal roots, if D = 0 Therefore, the value of (ii) The given quadric equation is , and roots are real and equal Then find the value of k. Here, So, As we know that Putting the value of The given equation will have real and equal roots, if D = 0 Now factorizing of the above equation So, either Therefore, the value of (iii) The given quadratic equation is ${x}^{2}-4kx+k=0$, and roots are real and equal. Then find the value of k. Here, ${x}^{2}-4kx+k=0$ So, $a=1,b=-4k\mathrm{and}c=k.$ As we know that $D={b}^{2}-4ac$ Putting the value of $a=1,b=-4k\mathrm{and}c=k.$ $D={\left(-4k\right)}^{2}-4\left(1\right)\left(k\right)\phantom{\rule{0ex}{0ex}}=16{k}^{2}-4k$ The given equation will have real and equal roots, if D = 0. So, $16{k}^{2}-4k=0$ Now factorizing the above equation, $16{k}^{2}-4k=0\phantom{\rule{0ex}{0ex}}⇒4k\left(4k-1\right)=0\phantom{\rule{0ex}{0ex}}⇒4k=0\mathrm{or}4k-1=0\phantom{\rule{0ex}{0ex}}⇒k=0\mathrm{or}k=\frac{1}{4}$ Therefore, the value of $k=0,\frac{1}{4}$. (iv) The given quadratic equation is $kx\left(x-2\sqrt{5}\right)+10=0$, and roots are real and equal. Then find the value of k. Here, $kx\left(x-2\sqrt{5}\right)+10=0\phantom{\rule{0ex}{0ex}}⇒k{x}^{2}-2\sqrt{5}kx+10=0$ So, $a=k,b=-2\sqrt{5}k\mathrm{and}c=10.$ As we know that $D={b}^{2}-4ac$ Putting the value of $a=k,b=-2\sqrt{5}k\mathrm{and}c=10.$ $D={\left(-2\sqrt{5}k\right)}^{2}-4\left(k\right)\left(10\right)\phantom{\rule{0ex}{0ex}}=20{k}^{2}-40k$ The given equation will have real and equal roots, if D = 0. So, $20{k}^{2}-40k=0$ Now factorizing the above equation, $20{k}^{2}-40k=0\phantom{\rule{0ex}{0ex}}⇒20k\left(k-2\right)=0\phantom{\rule{0ex}{0ex}}⇒20k=0\mathrm{or}k-2=0\phantom{\rule{0ex}{0ex}}⇒k=0\mathrm{or}k=2$ Therefore, the value of $k=0,2$. (v) The given quadratic equation is $px\left(x-3\right)+9=0$, and roots are real and equal. Then find the value of p. Here, $px\left(x-3\right)+9=0\phantom{\rule{0ex}{0ex}}⇒p{x}^{2}-3px+9=0$ So, $a=p,b=-3p\mathrm{and}c=9.$ As we know that $D={b}^{2}-4ac$ Putting the value of $a=p,b=-3p\mathrm{and}c=9.$ $D={\left(-3p\right)}^{2}-4\left(p\right)\left(9\right)\phantom{\rule{0ex}{0ex}}=9{p}^{2}-36p$ The given equation will have real and equal roots, if D = 0. So, $9{p}^{2}-36p=0$ Now factorizing the above equation, $9{p}^{2}-36p=0\phantom{\rule{0ex}{0ex}}⇒9p\left(p-4\right)=0\phantom{\rule{0ex}{0ex}}⇒9p=0\mathrm{or}p-4=0\phantom{\rule{0ex}{0ex}}⇒p=0\mathrm{or}p=4$ Therefore, the value of $p=0,4.$ (vi) The given quadratic equation is $4{x}^{2}+px+3=0$, and roots are real and equal. Then find the value of p. Here, $4{x}^{2}+px+3=0$ So, $a=4,b=p\mathrm{and}c=3.$ As we know that $D={b}^{2}-4ac$ Putting the value of $a=4,b=p\mathrm{and}c=3.$ $D={\left(p\right)}^{2}-4\left(4\right)\left(3\right)\phantom{\rule{0ex}{0ex}}={p}^{2}-48$ The given equation will have real and equal roots, if D = 0. So, ${p}^{2}-48=0$ Now factorizing the above equation, ${p}^{2}-48=0\phantom{\rule{0ex}{0ex}}⇒{p}^{2}-{\left(4\sqrt{3}\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\left(p-4\sqrt{3}\right)\left(p+4\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒p-4\sqrt{3}=0\mathrm{or}p+4\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒p=4\sqrt{3}\mathrm{or}p=-4\sqrt{3}$ Therefore, the value of $p=±4\sqrt{3}.$

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